Question 14 Marks
Verify that $2$ is a zero of the polynomial $x^3+4 x^2-3 x-18$.
AnswerLet $p(x)=x^3+4 x^2-3 x-18$
Now, $p(2)=2^3+4 \times 2^2-3 \times 2-18=0$
$\therefore$ $2$ is a zero of $p(x).$
View full question & answer→Question 24 Marks
If $1$ and $-2$ are two zeros of the polynomial $(x^3-4 x^2-7 x+10)$, find its third zero.
AnswerLet $f(x)=x^3-4 x^2-7 x+10$
Since 1 and -2 are the zeros of $f(x)$, it follows that each one of $(x-1)$ and $(x+2)$ is a factor of $f(x)$.
Consequently, $(x-1)(x+2)=\left(x^2+x-2\right)$ is a factor of $f(x)$.
On dividing $f(x)$ by $\left(x^2+x-2\right)$, we get:
$\therefore f(x) = 0$
$\Rightarrow (x^2+ x - 2)(x - 5) = 0$
$\Rightarrow (x - 1)(x + 2)(x - 5) = 0$
$\Rightarrow x = 1$ or $x = -2$ or $x = 5$
Hence, the third zero is $5$ View full question & answer→Question 34 Marks
Show that $(x + 2)$ is a factor of $f(x) = x^3+4 x^2+x-6 $.
Answer$ \text { Given: } f(x)=x^3+4 x^2+x-6 $
$ \text { Now, } f(-2)=(-2)^3+4(-2)^2+(-2)-6 $
$ =-8+16-2-6 $
$ =0$
$\therefore(\mathrm{x}+2) \text { is a factor of } \mathrm{f}(\mathrm{x})=\mathrm{x}^3+4 \mathrm{x}^2+\mathrm{x}-6 \text {. }$
View full question & answer→Question 44 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$3x^2- x - 4$
AnswerWe have,
$ f(x)=3 x^2-x-4$
$=3 x 2-4 x+3 x-4$
$=x(3 x-4)+1(3 x-4)$
$=(3 x-4)(x+1)$
$\therefore f(x)=0$
$\Rightarrow(3 x-4)(x+1)=0$
$\Rightarrow 3 x-4=0 \text { or } x+1=0$
$\Rightarrow x=\frac{4}{3} \text { or } x=-1$
So the zeros of $f(x)$ are $\frac{4}3{}$ and $-1$
Sum of zeros $=\frac{4}{3}+(-1)$
$=\frac{1}{3}=\frac{-(\text{Coefficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $=\frac{4}{3}\times(-1)$
$=\frac{-4}{3}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 54 Marks
Find all the zeros of $(2 x^4-3 x^3-5 x^2+9 x-3),$ if it is given that two of its zeros are $\sqrt3$ and $-\sqrt3.$
AnswerLet $f(x) = 2 x^4-3 x^3-5 x^2+9 x-3$
Since $\sqrt3$ and $-\sqrt3$ are the zeros of $f(x)$, it follows that each one of $\big(\text{x}-\sqrt3\big)$ and $(\text{x}+\sqrt3)$ is a factor of f(x).
Consequently, $\big(\text{x}-\sqrt3\big)\big(\text{x}+\sqrt3\big)=(\text{x}^2-3)$ is a factor of $f(x).$
On dividing f(x) by $(x^2- 3),$ we get:
$ \therefore f(x)=0$
$ \Rightarrow 2 x^4-3 x^3-5 x^2+9 x-3=0 $
$ \Rightarrow\left(x^2-3\right)\left(2 x^2-3 x+1\right)=0 $
$ \Rightarrow\left(x^2-3\right)\left(2 x^2-2 x-x+1\right)=0$
$\Rightarrow\big(\text{x}-\sqrt3\big)\big(\text{x}+\sqrt3\big)(\text{2x}+1)(\text{x}-1)=0$
$\Rightarrow\text{x}=\sqrt3$ or $\text{x}=-\sqrt3$ or $\text{x}=\frac{1}{2}$ or $x = 1$ View full question & answer→Question 64 Marks
Verify that $5, -2$ and $\frac{1}3{}$ are the zeros of the cubic polynomial $p(x) = 3 x^3-10 x^2-27 x+10$ and verify the relation between its zeros and coefficients.
Answer$p(x)=\left(3 x^3-10 x^2-27 x+10\right) $
$p(5)=\left(3 \times 5^3-10 \times 5^2-27 \times 5+10\right) $
$=(375-250-135+10)=0 $
$p(-2)=\left[3 \times\left(-2^3\right)-10 \times\left(-2^2\right)-27 \times(-2)+10\right] $
$=(-24-40+54+10)=0$
$\text{p}\Big(\frac{1}{3}\Big)=\Big\{3\times\Big(\frac{1}{3}\Big)^3-10\times\Big(\frac{1}3{}\Big)^2-27\times\frac{1}{3}+10\Big\}$
$=\Big(3\times\frac{1}{27}-10\times\frac{1}{9}-9+10\Big)$
$=\Big(\frac{1}{9}-\frac{10}{9}+1\Big)=\Big(\frac{1-10+9}{9}\Big)$
$=\Big(\frac{0}{9}\Big)=0$
$\therefore$ 5, -2 and $\frac{1}{3}$ are the zeroes of p(x),
Let $\alpha=5,\ \beta=-2$ and $\gamma=\frac{1}{3}.$ Then we have:
$(\alpha+\beta+\gamma)=\Big(5-2+\frac{1}{3}\Big)$
$=\Big(\frac{10}{3}\Big)=\frac{-(\text{Coefficient of }\text{x}^2)}{(\text{Coefficient of }\text{x}^3)}$
$(\alpha\beta+\beta\gamma+\gamma\alpha)=\Big(-10-\frac{2}{3}+\frac{5}{3}\Big)$
$=\frac{-27}{3}=\frac{\text{Coefficient of x}}{\text{Coefficient of }\text{x}^3}$
$\alpha\beta\gamma=\Big\{5\times(-2)\times\frac{1}{3}\Big\}$
$=\frac{-10}{3}=\frac{-\text{(Constant term)}}{(\text{coefficient of }\text{x}^3)}$
View full question & answer→Question 74 Marks
Very-Short-Answer Question:
The sum of the zero and the product of zero of a quadratic polynomial are $\frac{-1}{2}$ and $−3$ respectively, write the polynomial.
AnswerLet $\alpha$ and $\beta$ be the zeros of the required quadratic polynomial.
Then, we have
$\alpha+\beta=-\frac{1}{2}$ and $\alpha\beta=-3$
Now, a qudratic polynomial whose zeros are $\alpha$ and $\beta$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\therefore$ Required quadratic polynomial,
$\text{p}(\text{x})=\text{x}^2-\Big(-\frac{1}{2}\Big)\text{x}+(-3)$
$=\text{x}^2+\frac{1}{2}\text{x}-3$
View full question & answer→Question 84 Marks
Find the quadratic polynomial whose zeros are $2$ and $-6$. Verify the relation between the coeficients and the zeros of the polynomial.
AnswerLet $\alpha =2$ and $\beta=-6$
Sum of the zeros $=(\alpha+\beta)$
$=2+(-6) =- 4$
Product of the zeros $=\alpha\beta$
$=2\times(-6) =- 12$
$\therefore$ Required polynomial $=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-(-4)\text{x}-12$
$=\text{x}^2+\text{4x}-12$
Sum of the zeros = -4
$=\frac{-4}{1}=\frac{-(\text{Coefficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of the zeros = -12
$=\frac{-12}{1}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 94 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$2 x^2-11 x+15 x$
AnswerWe have,
$f(x)=2 x^2-11 x+15 x$
$ =2 x^2-6 x-5 x+15$
$ =2 x(x-3)-5(x-3)$
$ =(x-3)(2 x-5)$
$ \therefore f(x)=0$
$ \Rightarrow(x-3)(2 x-5)=0$
$ \Rightarrow x-3=0 \text { or } 2 x-5=0$
$ \Rightarrow x=3 \text { or } x=\frac{5}{2}$
So, the zeros of f(x) are 3 and $\frac{5}{2}$
Sum of zeros $3+\frac{5}{2}=\frac{11}{2}$
$=-\frac{(-11)}{2} =\frac{-(\text{Cofficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $=3\times\frac{5}{2}$
$=\frac{15}{2}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 104 Marks
Find a cubic polynomial whose zeroes are $2, -3$ and $4$
AnswerIf the zeroes of the cubic polynomial are $a, b$ and $c$ then the cubic polynomial can be found as:
$x^3-(a+b+c) x^2+(a b+b c+c a) x-a b c \ldots(1)$
Let $a=2, b=-3$ and $c=4$
Substituting the values in $(1)$, we get
$ x^3-(2-3+4) x^2+(-6-12+8) x-(-24)$
$ \Rightarrow x^3-3 x^2-10 x+24$
View full question & answer→Question 114 Marks
If 2 and $-2$ are two zeros of the polynomial $(x^4+x^3-34 x^2-4 x+120),$ find all the zeros of given polynomial.
AnswerLet $f(x)=x^4+x^3-34 x^2-4 x+120$
Since $2$ and $-2$ are the zeros of $f(x)$, it follows that each one of $(x-2)$ and $(x+2)$ is a factor of $f(x)$.
Consequently, $(x-2)(x+2)=\left(x^2-4\right)$ is a factor of $f(x)$.
On dividing $f(x)$ by $\left(x^2-4\right)$, we get:
$ \therefore f(x)=0 $
$ \Rightarrow\left(x^2+x-30\right)\left(x^2-4\right)=0 $
$ \Rightarrow\left(x^2+6 x-5 x-30\right)(x-2)(x+2)=0 $
$ \Rightarrow[x(x+6)-5(x+6)](x-2)(x+2)=0 $
$ \Rightarrow(x-5)(x+6)(x-2)(x+2)=0 $
$ \Rightarrow x=5 \text { or } x=-6 \text { or } x=2 \text { or } x=-2$
Hence, all the zeros are $2, -2, 5$ and $-6$ View full question & answer→Question 124 Marks
One zero of the polynomial $3x^3 + 16x^2 + 15x - 18$ is $\frac{2}{3}.$ Find the other zeros of the polynomial.
Answer$p(x) = 3x^3 + 16x^2 + 15x - 18$
Since $\frac{2}{3}$ is a zero of p(x), so $\Big(\text{x}-\frac{2}{3}\Big)$ is a factor of $f(x).$
$\Rightarrow (3x - 2)$ is also its factor
On dividing p(x) by $(3x - 2)$, we get
$ \therefore f(x)=\left(x^2+6 x+9\right)(3 x-2)$
$ =(x 2+3 x+3 x+9)(3 x-2)$
$ =[x(x+3)+3(x+3)](3 x-2)$
$ =(x+3)(x+3)(3 x-2)=0$
$ \therefore f(x)=0$
$ (x+3)(x+3)(3 x+2)=0$
$ x=-3 \text { or } x=-3 \text { or } x=\frac{2}{3}$
Thus, the other two zeros are $-3, -3$ View full question & answer→Question 134 Marks
It is given that $-1$ is one of the zeros of the polynomial $x^3 + 2x^2 - 11x - 12$. Find all the given zeros of the given polynomial.
AnswerLet $f(x) = x^3 + 2x^2 - 11x - 12$
Since $-1$ is a zero of $f(x), (x + 1)$ is a factor of $f(x).$
On dividing $f(x)$ by $(x + 1)$, we get:
$f(x)=x^3+2 x^2-11 x-12$
$=(x+1)\left(x^2+x-12\right)$
$=(x+1)\left\{x^2+4 x-3 x-12\right\}$
$=(x+1)\{x(x+4)-3(x+4)\}$
$=(x+1)(x-3)(x+4)$
$\therefore f(x)=0$
$\Rightarrow(x+1)(x-3)(x+4)=0$
$\Rightarrow(x+1)=0 \text { or }(x-3)=0 \text { or }(x+4)=0$
$\Rightarrow x=-1 \text { or } x=3 \text { or } x=-4$
Thus, all the zeros are $-1, 3$ and $-4$ View full question & answer→Question 144 Marks
Very-Short-Answer Question:
If $(a - b)$, a and $(a + b)$ are zeros of the polynomial $2 x^3-6 x^2+5 x-7$, write the value of a.
AnswerGiven polynomial is $p(x) = 2 x^3-6 x^2+5 x-7$
Let $\alpha=(\text{a}-\text{b}),\ \beta=\text{a}$ and $\gamma(\text{a}+\text{b})$
Now, $\alpha+\beta+\gamma =-\frac{(-6)}{2}=3$
$\Rightarrow (a - b) + a + (a + b) = 3$
$\Rightarrow 3a = 3$
$\Rightarrow a = 1$
View full question & answer→Question 154 Marks
If the zeroes of the polynomial $x^3-3 x^2+x+1$ are $(a - b)$, a and $(a + b)$, find the values of $a$ and $b.$
AnswerThe given polynomial $=x^3-3 x^2+x+1$ and its roots are $(a-b), a$ and $(a+b)$. Comparing the given polynomial with $\mathrm{Ax}^3+\mathrm{Bx}^2+\mathrm{Cx}+\mathrm{D}$,we have:
$A=1, B=-3, C=1$ and $D=1$
Now, $(\text{a}-\text{b})+\text{a}+(\text{a}+\text{b})=\frac{-\text{B}}{\text{A}}$
$\Rightarrow\text{3a}=-\frac{-3}{1}$
$\Rightarrow\text{a}=1$
Also, $(\text{a}-\text{b})\times\text{a}\times(\text{a}+\text{b})=\frac{-\text{D}}{\text{A}}$
$\Rightarrow\text{a}(\text{a}^2-\text{b}^2)=\frac{-1}{1}$
$\Rightarrow1(1^2-\text{b}^2)=-1$
$\Rightarrow1-\text{b}^2=-1$
$\Rightarrow\text{b}^2=2$
$\Rightarrow\text{b}=\pm\sqrt2$
$\therefore\text{a}=1$ and $\text{b}=\pm\sqrt2$
View full question & answer→Question 164 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$8x^2 - 4$
AnswerWe have,
$\text{f}(\text{x})=\text{8x}^2-4$
$=4(\text{2x}-1)^2$
$=4\big[\big(\sqrt2\text{x}\big)^2-1^2\big]$ $\big[\therefore\text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
$=4\big(\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+1\big)$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\big(\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+1\big)=0$
$\therefore\sqrt2\text{x}-1=0$ or $\sqrt2\text{x}+1=0$
$\therefore\text{x}=\frac{1}{\sqrt2}$ or $\text{x}=-\frac{1}{\sqrt2}$
So the zeros of $f(x)$ are $\frac{1}{\sqrt2}$ and $-\frac{1}{\sqrt2}$
Sum of zeros $=\Big(\frac{1}{\sqrt2}\Big)+\Big(-\frac{1}{\sqrt2}\Big)=0$
$=\frac{0}{8}=-\frac{\text{Coefficient of x}}{\text{Coefficient of }\text{x}^2}$
Product of zeros $=\Big(\frac{1}{\sqrt2}\Big)\times\Big(-\frac{1}{\sqrt2}\Big)=-\frac{1}{2}$
$=\frac{-4}{8}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 174 Marks
If $\alpha,\ \beta,\ \gamma$ are the zeroes of the polynomial $p(x)=6 x^3+3 x^2-5 x+1 $, find the value of $\Big(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\Big).$
Answer$ \text { Given: } p(x)=6 x^3+3 x^2-5 x+1 $
$ =6 x^2-(-3) x^2+(-5) x-(-1)$
Comparing the polynomial with $\text{x}^3-\text{x}^2(\alpha+\beta+\gamma)+\text{x}(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma,$ we get:
$\alpha\beta+\beta\gamma+\gamma\alpha=-5$
and $\alpha\beta\gamma=-1$
$\therefore\Big(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\Big)$
$=\Big(\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}\Big)$
$=\Big(\frac{-5}{-1}\Big)$
$=5$
View full question & answer→Question 184 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$4 x^2-4 x+1 $
AnswerWe have,
$ f(x)=4 x^2-4 x+1 $
$ =4 x^2-2 x-2 x+1 $
$ =2 x(2 x-1)-1(2 x-1) $
$ =(2 x-1)(2 x-1)$
$ \therefore$ $ f(x)=0 $
$ \Rightarrow(2 x-1)(2 x-1)=0 $
$ \Rightarrow(2 x-1)^2=0 $
$ \Rightarrow 2 x-1=0$
$\Rightarrow\text{x}=\frac{1}{2}$
So, the zeros of $f(x)$ are $\frac{1}{2}$ and $\frac{1}{2}$
Sum of zeros $\frac{1}{2}+\frac{1}{2}=1$
$=\frac{1}{1}=\frac{1\times4}{1\times4}$
$=\frac{4}{4} =\frac{-(\text{Cofficient of x})}{(\text{Coefficient of }\text{x}^2)}$
Product of zeros $=\frac{1}{2}\times\frac{1}{2}$
$=\frac{1}{4}=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
View full question & answer→Question 194 Marks
If $\text{x}=\frac{2}{3}$ and $x = -3$ are the roots of the quadratic equation $ax^2+ 7x + b = 0$ then find the values of a and b.
AnswerSince $\frac{2}{3}$ and $-3$ are zeros of $ax^2+ 7x + b,$ we have
Sum of roots $=\frac{2}{3}+(-3)$
$=\frac{2-9}{3}=\frac{-7}{3}$
$\Rightarrow\frac{-7}{\text{a}}=\frac{-7}{3}$
$\Rightarrow\text{a}=3$
Product of roots $=\frac{2}{3}\times(-3)=-2$
$\Rightarrow\frac{\text{b}}{\text{a}}=-2$
$\Rightarrow\frac{\text{b}}{3}=-2$
$\Rightarrow\text{b}=-6$
View full question & answer→Question 204 Marks
By actual division, show that $x^2- 3$ is a factor of $2 x^4+3 x^3-2 x^2-9 x-12$
AnswerLet $f(x) = 2 x^4+3 x^3-2 x^2-9 x-12$ and $g(x) = x^2- 3$
Quotient $\mathrm{q}(\mathrm{x})=2 \mathrm{x}^2+3 \mathrm{x}+4$
Remainder $r(x)=0$
Since, the remainder is $0 .$
Hence, $x^2-3$ is a factor of $2 x^4+3 x^3-2 x^2-9 x-12$ View full question & answer→Question 214 Marks
Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time and the product of its zeros are $5, -2$ and $-24$ respectively.
AnswerWe know the sum, sum of the product of the zeros taken two at a time and the product of the zeros of a cubic polynomial then the cubic polynomial can be found as:
$x^3-($ Sum of the zeros $) x^2+($ sum of the product of the zeros taking two at a time) $x$ - Product of zeros Therefore, the required polynomial is:
$x^3-5 x^2-2 x+24$
View full question & answer→Question 224 Marks
Find the quotient and the remainder when:
$f(x)=x^3-3 x^2+5 x-3$ is divided by $g(x)=x^2-2$
Answer
Quotient $q(x) = x - 3$
Remainder $r(x) = 7x - 9$ View full question & answer→Question 234 Marks
Find the quadratic polynomial, sum of whose zeros is $8$ and their product is $12$. Hence, find the zeros of the polynomial.
AnswerLet $\alpha,\beta$ be the zeros of required quadratic polynomial $f(x).$
Now, $\alpha+\beta=8$ and $\alpha\beta=12$
$\text{f}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-\text{8x}+12$
$\therefore$ required polynomial is $x^2-8 x+12$
$ \text { Also } f(x)=x^2-8 x+12 $
$ =x^2-6 x-2 x=12 $
$ =x(x-6)-2(x-6) $
$ \therefore f(x)=0 $
$ (x-6)(x-2)=0 $
$ \therefore x-6=0 \text { or } x-2=0 $
$ \text { i.e. } x=6 \text { or } x=2$
$\therefore$ Zeros of polynomial are $6$ and $2$
View full question & answer→Question 244 Marks
Verify division algorithm for the polynomials $f(x)=8+20 x+x^2-6 x^3$ and $g(x)=2+5 x-3 x^2$
Answer We can write $f(x)$ as $-6 x^3+x^2+20 x+8$ and $g(x)$ as $-3 x^2+5 x+2$
Quotient $= 2x + 3$
Remainder $= x + 2$
By using division rule, we have
Divided = Quotient × Divisor + Remainder
$\therefore-6 x^3+x^2+20 x+8=\left(-3 x^2+5 x+2\right)(2 x+3)+x+2 $
$ \Rightarrow-6 x^3+x^2+20 x+8=-6 x^3+10 x^2+4 x-9 x^2+15 x+6+x+2$
$ \Rightarrow-6 x^3+x^2+20 x+8=-6 x^3+x^2+20 x+8$ View full question & answer→Question 254 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$5y^2+ 10y$
AnswerWe have,
$f(x) = 5y^2+ 10y$
$=5y(y + 2)$
$\therefore f(x) = 0$
$⇒ 5y(y + 2) = 0$
$⇒ 5y = 0 or y + 2 = 0$
$⇒ y = 0 or y = -2$
So the zeros of $f(x)$ are $0$ and $-2$
Sum of zeros $= 0 + (-2) = -2$
$=\frac{-2}{1}=\frac{-2\times5}{1\times5}$
$=\frac{-10}{5}=-\frac{\text{Coefficient of y}}{\text{Coefficient of }\text{y}^2}$
Product of zeros $= 0 × (-2) = 0$
$=\frac{0}{1}=\frac{0\times5}{1\times5}$
$=\frac{0}{5}=\frac{\text{Constant term}}{\text{Coefficient of }\text{y}^2}$
View full question & answer→Question 264 Marks
Find the quotient and the remainder when:
$f(x)=x^4-5 x+6$ is divided by $g(x)=2-x^2$
AnswerWe can write
$f(x)$ as $x^4+0 x^3+0 x^2-5 x+6$ and $g(x)$ as $-x^2+2$
Quotient $q(x) =-x^2+2$
Remainder $r(x) = -5x + 10$ View full question & answer→Question 274 Marks
Find the quadratic polynomial, the sum of whose zeroes is $-5 $and their product is $6.$
AnswerGiven:
Sum of the zeroes$ = -5$
Product of the zeroes$ = 6$
$\therefore \text { Required polynomial }=\mathrm{x}^2-(\text { sum of the zeroes }) \mathrm{x}+\text { product of the zeros }$
$ =x^2-(-5) x+6 $
$ =x^2+5 x+6$
View full question & answer→Question 284 Marks
Show that the polynomial $f(x) = x^4+ 4x^2+ 6$ has no zeroes.
AnswerLet $t=x^2$
So, $f(t)=t^4+4 t^2+6$
Now, to find the zeros, we will equate $f(t)=0$
$\Rightarrow t^4+4 t^2+6=0$
Now, $\text{t}=\frac{-4\pm\sqrt{16-24}}{2}$
$=\frac{-4\pm\sqrt{-8}}{2}$
$=2\pm\sqrt{-2}$
i.e., $\text{x}^2=-2\pm\sqrt{-2}$
$\Rightarrow\text{x}=\sqrt{-2\pm\sqrt{-2}},$ which is not a real number.
The zeros of a polynomial should be real number s.
$\therefore$ The given $f(x)$ has no zeros.
View full question & answer→Question 294 Marks
Verity that $3, -2, 1$ are the zeros of the cubic polynomial $ p(x)=\left(x^3-2 x^2-5 x+6\right)$ and verify the relation between its zeros and coefficients.
Answer$\text { The given polynomial is } p(x)=\left(x^3-2 x^2-5 x+6\right)$
$ p(3)=\left(3^3-2 \times 3^2-5 \times 3+6\right) $
$ =(27-18-15+6)=0 $
$ p(-2)=\left[\left(-2^3\right)-2 \times\left(-2^2\right)-5 \times(-2)+6\right] $
$ =(-8-8+10+6)=0 $
$ p(1)=\left(1^3-2 \times 1^2-5 \times 1+6\right)=(1-2-5+6)=0$
$3, -2$ and $1$ are the zeroes of $p(x),$
Let $\alpha=3,\ \beta=-2$ and $\gamma=1.$ Then we have:
$(\alpha+\beta+\gamma)=(3-2+1)$
$=2=\frac{-(\text{Coefficient of }\text{x}^2)}{(\text{Coefficient of }\text{x}^3)}$
$(\alpha\beta+\beta\gamma+\gamma\alpha)=(-6-2+3)$
$=\frac{-5}{1}=\frac{\text{Coefficient of x}}{\text{Coefficient of }\text{x}^3}$
$\alpha\beta\gamma=\Big\{3\times(-2)\times1\Big\}$
$=\frac{-6}{1}=\frac{-\text{(Constant term)}}{(\text{coefficient of }\text{x}^3)}$
View full question & answer→Question 304 Marks
Short-Answer Question:
If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x) = 6x^2+ x - 2,$ find the value of $\Big(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\Big).$
AnswerSince $\alpha$ and $\beta$ are the zeros of $6x^2+ x - 2,$ we have
$\alpha+\beta=-\frac{1}{6}$
$\alpha\beta=\frac{-2}{6}=\frac{-1}{3}$
$\therefore\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}$
$=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$
$=\frac{\Big(-\frac{1}{6}\Big)^2-2\Big(-\frac{1}{3}\Big)}{\Big(-\frac{1}{3}\Big)}$
$=-\frac{\frac{1}{36}+\frac{2}{3}}{\frac{1}{3}}$
$=-\frac{\frac{25}{36}}{\frac{1}{3}}$
$=-\frac{25}{36}\times\frac{3}{1}$
$=-\frac{25}{12}$
View full question & answer→Question 314 Marks
Short-Answer Question:
If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x) = x^2- 5x + k,$ such that $\alpha-\beta=1$ find the value of k.
Answer Let are the zeros of $x^2- 5x + k.$
Then, we have
$\alpha+\beta=-\frac{(-5)}{1}=5$
$\alpha\beta=\frac{\text{k}}1{}=\text{k}$
Given, $\alpha-\beta=1$
Now, $(\alpha+\beta)^2=(\alpha-\beta)^2+4\alpha\beta$
$\Rightarrow(5)^2=(1)^2+4\times\text{k}$
$\Rightarrow25=1+\text{4k}$
$\Rightarrow\text{4k}=24$
$\Rightarrow\text{k}=\frac{24}{4}$
$\Rightarrow\text{k}=6$
View full question & answer→Question 324 Marks
If $(x + a)$ is a factor of the polynomial $2 x^2+2 a x+5 x+10$, find the value of of a.
AnswerLet $f(x)=2 x^2+2 a x+5 x+10$
Since $(x+a)$ is a factor of $f(x)$, we have
$ f(-a)=0 $
$ \Rightarrow 2(-a)^2+2 a(-a)+5(-a)+10=0 $
$ \Rightarrow 2 a^2-2 a^2-5 a+10=0 $
$ \Rightarrow 5 a=10 $
$ \Rightarrow a=2$
View full question & answer→Question 334 Marks
Short-Answer Question:
If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x) = 5x^2- 7x + 1,$ find the value of $\Big(\frac{1}{\alpha}+\frac{1}{\beta}\Big).$
AnswerSince $\alpha$ and $\beta$ are the zeros of $5x^2- 7x + 1,$ we have
$\alpha+\beta=-\frac{(-7)}{5}=\frac{7}{5}$
$\alpha\beta=\frac{1}{5}$
$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha\beta}$
$=\frac{\frac{7}{5}}{\frac{1}{5}}$
$=\frac{7}{5}\times\frac{5}{1}$
$=7$
View full question & answer→Question 344 Marks
Short-Answer Question:
If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x) = x^2+ x - 2,$ find the value of $\Big(\frac{1}{\alpha}-\frac{1}{\beta}\Big).$
AnswerSince $\alpha$ and $\beta$ are the zeros of $x^2+ x - 2,$ we have
$\alpha+\beta=-\frac{1}{1}=-1$
$\alpha\beta=\frac{-2}{1}=-2$
$\therefore\frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta}$
$=\frac{\sqrt{(\beta+\alpha)^2-4\alpha\beta}}{\alpha\beta}$ $\dots[(\text{a}-\text{b})^2=(\text{a}+\text{b})^2-\text{4ab}]$
$=\frac{\sqrt{(-1)^2-4(-2)}}{-2}$
$=-\frac{\sqrt{1+8}}{2}$
$=-\frac{\sqrt9}{2}$
$=-\frac{3}{2}$
View full question & answer→Question 354 Marks
Find the quotient and the remainder when:
$f(x)=x^4-3 x^2+4 x+5$ is divided by $g(x)=x^2+1-x$
Answer
Quotient $q(x) = x^2 + x - 3$
Remainder $r(x) = 8$ View full question & answer→Question 364 Marks
Find the quadratic polynomial whose zeros are $\frac{2}{3}$ and $\frac{-1}{4}.$ Verify the relation between the coefficients and the zeros of the polynomial.
AnswerLet $\alpha$ and $\beta$ are the zeros then
$\alpha +\beta=\frac{2}{3}+\Big(-\frac{1}{4}\Big)$
$=\frac{8-3}{12}=\frac{5}{12}$
$\alpha\beta=\frac{2}{3}\times\Big(-\frac{1}{4}\Big)$
$=-\frac{2}{12}=-\frac{1}{6}$
$\therefore$ quadratic polynomial whose zeros are $\alpha,\beta$ is:
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-\Big(\frac{5}{12}\Big)\text{x}+\Big(-\frac{1}{6}\Big)$
$=\frac{1}{12}(\text{12x}^2-\text{5x}-2)$
Sum of zeros $=-\frac{\text{Coefficient of x}}{\text{Coefficients of }\text{x}^2}$
$=-\frac{-5}{12}=\frac{5}{12}$
Also sum of zeros $=\frac{2}{3}+\Big(-\frac{1}{4}\Big)$
$=\frac{5}{12}$
Product of zeros $=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
$=\frac{-2}{12}=\frac{-1}{6}$
Also product of zeros $=\frac{2}{3}\times\Big(-\frac{1}{4}\Big)$
$=\frac{-2}{12}=\frac{-1}{6}$
View full question & answer→Question 374 Marks
If two zeroes of the polynomial $p(x) = 2 x^4-3 x^3-3 x^2+6 x-2$ are $\sqrt2$ and $-\sqrt2,$ find its other two zeroes.
AnswerGiven: $p(x) = 2 x^4-3 x^3-3 x^2+6 x-2$ and the two zeros $\sqrt2$ and $-\sqrt2$
So, the polynomial is $\big(\text{x}+\sqrt2\big)\big(\text{x}-\sqrt2\big)=\text{x}^2-2$
Let us divide $p(x)$ by $(x^2- 2).$
$ \text { Here, } 2 x^4-3 x^3-3 x^2+6 x-2$
$ =\left(x^2-2\right)\left(2 x^2-3 x+1\right)$
$ =\left(x^2-2\right)\left[2 x^2-(2+1) x+1\right]$
$ =\left(x^2-2\right)\left(2 x^2-2 x-x+1\right)$
$ =\left(x^2-2\right)[(2 x(x-1)-1(x-1))]$
$ =\left(x^2-2\right)(2 x-1)(x-1)$
$\therefore$ The other two zeros are $\frac{1}{2}$ and $1.$
View full question & answer→Question 384 Marks
Short-Answer Question:
If the zeroes of a polynomial $f(x)=x^3-3 x^2+x+1$ are $(a - b)$, a and $(a + b)$, find a and $b.$
AnswerGiven Polynomial is $p(x)=x^3-3 x^2+x+1$
Let $\alpha=(\text{a}-\text{b}) ,\ \beta=\text{a}$ and $\gamma=(\text{a}+\text{b})$
Now, $\alpha+\beta+\gamma=-\frac{(-3)}{1}$
$\Rightarrow (a - b) + a + (a + b) = 3$
$\Rightarrow 3a = 3$
$\Rightarrow a = 1$
Also, $\alpha\beta+\beta\gamma+\gamma\alpha=\frac{1}{1}$
$\Rightarrow (a - b)a + a(a + b) + (a + b)(a - b) = 1$
$\Rightarrow a^2-a b+a^2+a b+a^2-b^2=1$
$\Rightarrow 3 a^2-b^2=1$
$\Rightarrow 3(1)^2-b^2=1$
$\Rightarrow \mathrm{b}^2=2$
$\Rightarrow\text{b}=\pm\sqrt2$
Hence, a = 1 and $\text{b}=\pm\sqrt2$
View full question & answer→Question 394 Marks
On dividing, $3 x^3+x^2+2 x+5$ by a polynomial g(x), the quotient and remainder are $3x - 5$ and $9x + 10$ respectively. Find $g(x)$
Hint: $\text{g}(\text{x})=\frac{(\text{3x}^3+\text{x}^2+\text{2x}+5)-(\text{9x}+10)}{(\text{3x}-5)}$
AnswerBy using division rule, we have
Divided = Quotient × Divisor + Remainder
$ \therefore 3 x^3+x^2+2 x+5=(3 x-5) g(x)+9 x+10 $
$ \Rightarrow 3 x^3+x^2+2 x+5-9 x-10=(3 x-5) g(x) $
$\Rightarrow 3 x^3+x^2-7 x-5=(3 x-5) g(x)$
$\Rightarrow\text{g}(\text{x})=\frac{3\text{x}^3+\text{x}^2-\text{7x}-5}{\text{3x}-5}$
$\therefore g(x) = x^2 + 2x + 1$ View full question & answer→Question 404 Marks
Find a cubic polynomial whose zeros are $3, 5$ and $-2$
AnswerLet $\alpha,\ \beta$ and $\gamma$ be the zeros of the required polynomial.
Then we have:
$\alpha+\beta+\gamma$
$=3+5+(-2)=6$
$\alpha\beta+\beta\gamma+\gamma\alpha$
$=3\times5+5\times(-2)+(-2)\times3=-1$
and $\alpha\beta\gamma=3\times5\times-2=-30$
Now, $\text{p}(\text{x})=\text{x}^3-\text{x}^2(\alpha+\beta+\gamma)+\text{x}(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma$
$=\text{x}^3-\text{x}^2\times6+\text{x}\times(-1)-(- 30)$
$=\text{x}^3-\text{6x}^2-\text{x}=30$
So, the required polynomial is $p(x)=x^3-6 x^2-x+30$
View full question & answer→Question 414 Marks
Find the quotient when $p(x)=3 x^4+5 x^3-7 x^2+2 x+2$ is divided by $\left(x^2+3 x+1\right)$.
AnswerGiven: $p(x)=3 x^4+5 x^3-7 x^2+2 x+2$
Dividing p(x) by $\left(x^2+3 x+1\right)$, we have:
$\therefore$ The quotient is $3x^2- 4x + 2$ View full question & answer→Question 424 Marks
Short-Answer Question:
Write the zeros of the quadratic polynomial $\text{f}(\text{x})=4\sqrt3\text{x}^2+\text{5x}-2\sqrt3$
AnswerWe have,
$\text{f}(\text{x})=4\sqrt3\text{x}^2+\text{5x}-2\sqrt3$
$=4\sqrt3\text{x}^2+\text{8x}-\text{3x}-2\sqrt3$
$=\text{4x}\big(\sqrt3\text{x}+2\big)-\sqrt3\big(\sqrt3\text{x}+2\big)$
$=\big(\text{4x}-\sqrt3\big)\big(\sqrt3\text{x}+2\big)$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\big(\text{4x}-\sqrt3\big)\big(\sqrt3\text{x}+2\big)=0$
$\Rightarrow\text{4x}-\sqrt3=0$ or $\sqrt3\text{x}+2$
$\Rightarrow\text{x}=\frac{\sqrt3}{4}$ or $\text{x}=-\frac{2}{\sqrt3}$
So, the zeros of $f(x) $are $\frac{\sqrt3}{4}$ and $-\frac{2}{\sqrt3}$
View full question & answer→Question 434 Marks
Find all the zeros of the polynomial $(2 x^4-11 x^3+7 x^2+13)$, it being given that two if its zeros are $3+\sqrt2$ and $3-\sqrt2.$V
AnswerLet $f(x) =2 x^4-11 x^3+7 x^2+13 x-7 $
Since $\big(3+\sqrt2\big)$ and $\big(3-\sqrt2\big)$ are the zeros of f(x), it follows that each one of $\big(\text{x}+3+\sqrt2\big)$ and $\big(\text{x}+3-\sqrt2\big)$ is a factor of f(x).
Consequently, $\big[\text{x}-\big(3+\sqrt2\big)\big]\big[\text{x}-\big(3-\sqrt2\big)\big]$
$=\big[(\text{x}-3)-\sqrt2\big]\big[(\text{x}-3)+\sqrt2\big]$
$=\big[(\text{x}-3)^2-2\big]=\text{x}^2-6\text{x}+7,$ which is a factor of $f(x).$
On dividing f(x) by $(x^2- 6x + 7),$ we get:
$ \therefore f(x)=0 $
$ \Rightarrow 2 x^4-11 x^3+7 x^2+13 x-7=0$
$ \Rightarrow\left(x^2-6 x+7\right)\left(2 x^2+x-1\right)=0$
$\Rightarrow\big(\text{x}+3+\sqrt2\big)\big(\text{x}+3-\sqrt2\big)\$\text{2x}-1)(\text{x}+1)=0$
$\Rightarrow\text{x}=-3-\sqrt2$ or $\text{x}=-3+\sqrt2$ or $\text{x}=\frac{1}{2}$ or $x = -1$
Hence, all the zeros are $\big(-3-\sqrt2\big),\ \big(-3+\sqrt2\big),\ \frac{1}{2}$ and $-1$ View full question & answer→Question 444 Marks
If 3 and -3 are two zeros of the polynomial $\left(x^4+x^3-11 x^2-9 x+18\right)$, find all the zeros of the given polynomial.
AnswerLet $f(x)=x^4+x^3-11 x^2-9 x+18$
Since $3$ and $-3$ are the zeros of $f(x),$ it follows that each one of $(x + 3)$ and $(x - 3) $is a factor of $f(x).$
Consequently, $(x-3)(x+3)=\left(x^2-9\right)$ is a factor of $f(x)$.
On dividing $f(x)$ by $\left(x^2-9\right)$, we get:
$ \therefore f(x)=0$
$ \Rightarrow\left(x^2+x-2\right)\left(x^2-9\right)=0 $
$ \Rightarrow\left(x^2+2 x-x-2\right)(x-3)(x+3)=0 $
$\Rightarrow(x-1)(x+2)(x-3)(x+3)=0$
$\Rightarrow x=1 \text { or } x=-2 \text { or } x=3 \text { or } x=-3$
Hence, all the zeros are $1, -2, 3$ and $-3$ View full question & answer→Question 454 Marks
If one zero of the polynomial $p(x) = x^3-6 x^2+11 x-6$ is $3,$ find the other two zeroes.
AnswerGiven: $p(x)=x^3-6 x^2+11 x-6$ and its factor, $x+3$
Let us divided $p(x)$ by $(x-3)$.
Here, $x^3-6 x^2+11 x-6$
$=(x-3)\left(x^2-3 x+2\right)$
$=(x-3)\left[x^2-(2+1) x+2\right]$
$=(x-3)\left(x^2-2 x-x+2\right)$
$=(x-3)[x(x-2)-1(x-2)]$
$=(x-3)(x-1)(x-2)$
$\therefore$ The other two zeros are $1$ and $2.$
View full question & answer→Question 464 Marks
Find a cubic polynomial whose zeroes are $\frac{1}{2}, 1$ and $−3.$
AnswerIf the zeroes of the cubic polynomial are $a, b$ and $c$ then the cubic polynomial can be found as:
$x^3- (a + b + c)x^2+ (ab + bc + ca)x - abc ...(1)$
Let $\text{a}=\frac{1}{2}, b = 1$ and $c = -3$
Substituting the values in $(1),$ we get
$\text{x}^3-\Big(\frac{1}{2}+1-3\Big)\text{x}^2+\Big(\frac{1}{2}-3-\frac{3}{2}\Big)\text{x}-\Big(\frac{-3}{2}\Big)$
$\Rightarrow\text{x}^3-\Big(\frac{-3}{2}\Big)\text{x}^2-\text{4x}+\frac{3}{2}$
$\Rightarrow\text{2x}^3+\text{3x}^2-\text{8x}+3$
View full question & answer→Question 474 Marks
Obtain all other zeros of $(x^4 + 4x^3- 2x^2- 20x - 15)$ if two of its zeros are $\sqrt5$ and $-\sqrt5.$
AnswerLet $f(x) = x^4+ 4x^3- 2x^2- 20x - 15$
Since $\sqrt5$ and $-\sqrt5$ are the zeros of f(x), it follows that each one of $\big(\text{x}-\sqrt5\big)$ and $(\text{x}+\sqrt5)$ is a factor of $f(x).$
Consequently, $\big(\text{x}-\sqrt5\big)\big(\text{x}+\sqrt5\big)=(\text{x}^2-5)$ is a factor of f(x).
On dividing f(x) by $(x^2- 5),$ we get:
$\therefore f(x) = 0$
$⇒ x^4+ 4x^3- 2x^2- 20x - 15$
$⇒ (x^2- 5)(x^2- 4x + 3)$
$\Rightarrow\big(\text{x}-\sqrt5\big)\big(\text{x}+\sqrt5\big)(\text{x}+1)(\text{x}+3)=0$
$\Rightarrow\text{x}=\sqrt5$ or $\text{x}=-\sqrt5$ or $x = -1$ or $x = -3$
Hence, all the zeros are $\sqrt5,\ -\sqrt5,\ -1$ and $-3$ View full question & answer→Question 484 Marks
Short-Answer Question:
Write the zeros of the quadratic polynomial $f(x) = 6x^2- 3$
AnswerTo find the zeros of the quadratic polynomial we will equate $f(x)$ to $0$
$ f(x)=0 $
$ \Rightarrow 6 x^2-3=0 $
$ \Rightarrow 3\left(2 x^2-1\right)=0 $
$ \Rightarrow 2 x^2-1=0 $
$ \Rightarrow 2 x^2=1$
$\Rightarrow\text{x}^2=\frac{1}{2}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt2}$
Hence, the zeros of the quadratic polynomial $f(x) = 6x^2- 3$ are $\frac{1}{\sqrt2},\ -\frac{1}{\sqrt2}.$
View full question & answer→Question 494 Marks
Find all the zeros of $(x^4 + x^3 - 23x^2 - 3x + 60),$ if it is given that two of its zeros are $\sqrt3$ and $-\sqrt3.$
AnswerLet $f(x) = x^4 + x^3 - 23x^2 - 3x + 60$
Since $\sqrt3$ and $-\sqrt3$ are the zeros of $f(x),$ it follows that each one of $\big(\text{x}-\sqrt3\big)$ and $(\text{x}+\sqrt3)$ is a factor of $f(x).$
Consequently, $\big(\text{x}-\sqrt3\big)\big(\text{x}+\sqrt3\big)=(\text{x}^2-3)$ is a factor of $f(x).$
On dividing $f(x)$ by $(x^2- 3),$ we get:
$ \therefore f(x)=0 $
$ \Rightarrow\left(x^2+x-20\right)\left(x^2-3\right)=0 $
$ \Rightarrow\left(x^2+5 x-4 x-20\right)\left(x^2-3\right)=0 $
$ \Rightarrow[x(x+5)-4(x+5)]\left(x^2-3\right)=0$
$\Rightarrow(\text{x}-4)(\text{x}+5) \big(\text{x}-\sqrt3\big)\big(\text{x}+\sqrt3\big)=0$
$⇒ x = 4$ or $x = -5$ or $\text{x}=\sqrt3$ or $\text{x}=-\sqrt3$
Hence, all the zeros are $\sqrt3,\ -\sqrt3,\ 4$ and $-5$ View full question & answer→Question 504 Marks
Use remainder theorem to find the value of $k$, it being given that when $\mathrm{p}(\mathrm{x})=\mathrm{x}^3+2 \mathrm{x}^2+\mathrm{kx}+3$ is divided by $(x - 3),$ then the remainder is $21.$
AnswerLet ${p}({x})={x}^3+2 {x}^2+{kx}+3$
Now, ${p}(3)=(3)^3+2(3)^2+3 {k}+3$
$=27+18+3 k+3 $
$ =48+3 k$
It is given that the remainder is $21$
$ \therefore 3 k+48=21 $
$ \Rightarrow 3 k=-27 $
$ \Rightarrow k=-9$
View full question & answer→