Question 15 Marks
Verify that 2 is a zero of the polynomial $x^3 + 4x^2 - 3x - 18.$
AnswerLet $p(x) = x^3 + 4x^2 - 3x - 18$
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 25 Marks
If 1 and -2 are two zeros of the polynomial ($x^3 - 4x^2 - 7x + 10)$, 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 35 Marks
Show that (x + 2) is a factor of $f(x) = x^3 + 4x^2 + x - 6.$
AnswerGiven: $f(x) = x^3 + 4x^2 + x - 6$
$Now, f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6$
$= -8 + 16 - 2 - 6$
$= 0$
$\therefore$ (x + 2) is a factor of $f(x) = x^3 + 4x^2 + x - 6.$
View full question & answer→Question 45 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, $f(x)=3 x^2-x-4=3 x 2-4 x+3 x-4=x(3 x-4)+1(3 x-4)=(3 x-4)(x+1) \therefore f(x)=0 \Rightarrow(3 x-4)(x+1)=0$
$\Rightarrow$ $3 x-4=0$ or $x+1=0$
$\Rightarrow x=\frac{4}{3}$ 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 55 Marks
Find all the zeros of $\left(2 x^4-3 x^3-5 x^2+9 x-3\right)$, if it is given that two of its zeros are $\sqrt{3}$ and $-\sqrt{3}$.
AnswerLet $f(x)=2 x^4-3 x^3-5 x^2+9 x-3$ Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeros of $f(x)$,
it follows that each one of $(x-\sqrt{3})$ and $(x+\sqrt{3})$ is a factor of $f(x)$.
Consequently, $(x-\sqrt{3})(x+\sqrt{3})=\left(x^2-3\right)$ is a factor of $f(x)$. On dividing $f(x)$ by $\left(x^2\right. - 3)$, 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(x-\sqrt{3})(x+\sqrt{3})(2 x+1)(x-1)=0 $
$\Rightarrow x=\sqrt{3} \text { or } x=-\sqrt{3} \text { or } x=\frac{1}{2} \text { or } x=1$ View full question & answer→Question 65 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$p(x) = (3x^3 - 10x^2 - 27x + 10) p(5) = (3 \times 5^3- 10 \times 5^2 - 27 \times 5 + 10) = (375 - 250 - 135 + 10) = 0 p(-2) = [3 \times (-2^3) - 10 \times (-2^2) - 27 \times (-2) + 10] = (-24 - 40 + 54 + 10) = 0$
$\text{p}\Big(\frac{1}{3}\Big)=\Big\{3\times\Big(\frac{1}{3}\Big)^3-10\times\Big(\frac{1}3{}\Big)^2-27\times\frac{1}{3}+10\Big\}$
$=\Big(3\times\frac{1}{27}-10\times\frac{1}{9}-9+10\Big)$
$=\Big(\frac{1}{9}-\frac{10}{9}+1\Big)=\Big(\frac{1-10+9}{9}\Big)$
$=\Big(\frac{0}{9}\Big)=0$
$\therefore$ 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 75 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 85 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 95 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$2x^2 - 11x + 15x$
AnswerWe have,
$f(x) = 2x^2 - 11x + 15x$
$= 2x^2 - 6x - 5x + 15$
$= 2x(x - 3) - 5(x - 3)$
$= (x - 3)(2x - 5)$
$\therefore$ f(x) = 0
$\Rightarrow (x - 3)(2x - 5) = 0$
$\Rightarrow x - 3 = 0 or 2x - 5 = 0$
⇒ x = 3 or $\text{x}=\frac{5}{2}$
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 105 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:
$x^3- (a + b + c)x^2 + (ab + bc + ca)x - abc ...(1)$
Let a = 2, b = -3 and c = 4
Substituting the values in (1), we get
$x^3 - (2 - 3 + 4)x^2 + (-6 - 12 + 8)x - (-24)$
$\Rightarrow x^3 - 3x^2 - 10x + 24$
View full question & answer→Question 115 Marks
If 2 and -2 are two zeros of the polynomial$ (x^4 + x^3 - 34x^2 - 4x + 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:
$\therefore$ $f(x) = 0 \Rightarrow (x^2 + x - 30)(x^2 - 4) = 0+$
$\Rightarrow (x^2 + 6x - 5x - 30)(x - 2)(x + 2) = 0$
$\Rightarrow [x(x + 6) - 5(x + 6)](x - 2)(x + 2) = 0$
$\Rightarrow (x - 5)(x + 6)(x - 2)(x + 2) = 0$
$\Rightarrow x = 5 or x = -6 or x = 2 or x = -2$
Hence, all the zeros are 2, -2, 5 and -6 View full question & answer→Question 125 Marks
One zero of the polynomial $3 x^3+16 x^2+15 x-18$ 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).
⇒ (3x - 2) is also its factor On dividing p(x) by (3x - 2), we get
$\therefore$ $f(x) = (x^2 + 6x + 9)(3x - 2) = (x2 + 3x + 3x + 9)(3x - 2)$
$= [x(x + 3) + 3(x + 3)](3x - 2) = (x + 3)(x + 3)(3x - 2) = 0$
$\therefore$ f(x) = 0 (x + 3)(x + 3)(3x + 2) = 0 x = -3 or x
= -3 or $\text{x}=\frac{2}{3}$ Thus, the other two zeros are -3, -3 View full question & answer→Question 135 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:
$f(x) = x^3+ 2x^2- 11x - 12$
$= (x + 1)(x^2 + x - 12)$
$= (x + 1){x^2 + 4x - 3x - 12}$
$= (x + 1){x(x + 4) - 3(x + 4)}$
$= (x + 1)(x - 3)(x + 4)$
$\therefore$ f(x) = 0
⇒ (x + 1)(x - 3)(x + 4) = 0
⇒ (x + 1) = 0 or (x - 3) = 0 or (x + 4) = 0
⇒ x = -1 or x = 3 or x = -4
Thus, all the zeros are -1, 3 and -4 View full question & answer→Question 145 Marks
Very-Short-Answer Question:
If (a - b), a and (a + b) are zeros of the polynomial $2x^3 - 6x^2 + 5x - 7$, write the value of a.
AnswerGiven polynomial is $p(x) = 2x^3 - 6x^2 + 5x - 7$
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 155 Marks
If the zeroes of the polynomial $x^3^- 3x^2 + x + 1$ are (a - b), a and (a + b), find the values of a and b.
AnswerThe given polynomial $= x^3^- 3x^2 + x + 1# and its roots are (a - b), a and (a + b).
Comparing the given polynomial with $Ax^3 + Bx^2 + Cx + D,$ 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 165 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 175 Marks
If $\alpha,\ \beta,\ \gamma$ are the zeroes of the polynomial $p(x) = 6x^3 + 3x^2^- 5x + 1$, find the value of $\Big(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\Big).$
AnswerGiven: $p(x) = 6x^3 + 3x^2^- 5x + 1$
$= 6x^2 - (-3)x^2 + (-5)x - (-1)$
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 185 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$4x^2 - 4x + 1$
AnswerWe have,
$f(x) = 4x^2 - 4x + 1$
$= 4x^2 - 2x - 2x + 1$
$= 2x(2x - 1) - 1(2x - 1)$
$= (2x - 1)(2x - 1)$
$\therefore$ f(x) = 0
$\Rightarrow (2x - 1)(2x - 1) = 0$
$\Rightarrow (2x - 1)^2 = 0$
$\Rightarrow 2x - 1 = 0$
$\Rightarrow\text{x}=\frac{1}{2}$
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 195 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 205 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 $q ( x )=2 x ^2+3 x +4$
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 215 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:
$x^3- 5x^2- 2x + 24$
View full question & answer→Question 225 Marks
Find the quotient and the remainder when:
$f(x) = x^3 - 3x^2 + 5x - 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 235 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 $x^2 - 8x + 12$
Also $f(x) = x^2 - 8x + 12$
$= x^2- 6x - 2x = 12$
$= x(x - 6) - 2(x - 6)$
$\therefore$ f(x) = 0
(x - 6)(x - 2) = 0
$\therefore$ x - 6 = 0 or x - 2 = 0
i.e. x = 6 or x = 2
$\therefore$ Zeros of polynomial are 6 and 2
View full question & answer→Question 245 Marks
Verify division algorithm for the polynomials $f(x)=8+20 x+x^2-6 x^3$ and $g(x)=2+5 x-3 x^2$
AnswerWe 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$$ -6x^3 + x^2 + 20x + 8 = (-3x^2 + 5x + 2)(2x + 3) + x + 2$
$\Rightarrow -6x^3 + x^2 + 20x + 8 = -6x^3 + 10x^2 + 4x - 9x^2 + 15x + 6 + x + 2$
$ \Rightarrow -6x^3 + x^2 + 20x + 8 = -6x^3 + x^2 + 20x + 8$ View full question & answer→Question 255 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$5y^2 + 10y$
AnswerWe have, $f(x)=5 y^2+10 y=5 y(y+2) \therefore f(x)=0 \Rightarrow 5 y(y+2)=0 \Rightarrow 5 y=0$ or $y+2=0 \Rightarrow y=0$ 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 265 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 275 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
$\therefore$ Required polynomial $= x^2$ - (sum of the zeroes)x + product of the zeros
$= x^2- (-5)x + 6$
$= x^2 + 5x + 6$
View full question & answer→Question 285 Marks
Show that the polynomial $f(x)=x^4+4 x^2+6$ has no zeroes.
AnswerLet $t = x^2$
So, $f(t) = t^4 + 4t^2 + 6$
Now, to find the zeros, we will equate f(t) = 0
$\Rightarrow t^4 + 4t^2 + 6 = 0$
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 295 Marks
Verity that 3, -2, 1 are the zeros of the cubic polynomial $p(x) = x^3 - 2x^2 - 5x + 6$ and verify the relation between its zeros and coefficients.
AnswerThe given polynomial is $p(x) = (x^3 - 2x^2 - 5x + 6)$
$p(3) = (3^3- 2 \times 3^2 - 5 \times 3 + 6)$
$= (27 - 18 - 15 + 6) = 0$
$p(-2) = [(-2^3) - 2 \times (-2^2) - 5 \times (-2) + 6]$
$= (-8 - 8 + 10 + 6) = 0$
$p(1) = (1^3 - 2 \times 1^2 - 5 \times 1 + 6) = (1 - 2 - 5 + 6) = 0$
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 305 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 315 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.
AnswerLet 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 325 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
$f(-a) = 0$
$\Rightarrow 2(-a)^2 + 2a(-a) + 5(-a) + 10 = 0$
$\Rightarrow 2a^2 - 2a^2 - 5a + 10 = 0$
$\Rightarrow 5a = 10$
$\Rightarrow a = 2$
View full question & answer→Question 335 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 345 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 355 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 365 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 375 Marks
If two zeroes of the polynomial $p(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2$ 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 $\sqrt{2}$ and $-\sqrt{2}$ So, the polynomial is
$(x+\sqrt{2})(x-\sqrt{2})=x^2-2$
Let us divide $p(x)$ by $\left(x^2-2\right)$.
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 385 Marks
Short-Answer Question:
If the zeroes of a polynomial $f(x) = x^3 - 3x^2+ x + 1$ are (a - b), a and (a + b), find a and b.
AnswerGiven Polynomial is $p(x) = x^3 - 3x^2+ x + 1$
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, $\alpha\beta+\beta\gamma+\gamma\alpha=\frac{1}{1}$
$\Rightarrow (a - b)a + a(a + b) + (a + b)(a - b) = 1$
$\Rightarrow a^2 - ab + a^2 + ab + a^2 - b^2 = 1$
$\Rightarrow 3a^2- b^2 = 1$
$\Rightarrow 3(1)^2 - b^2 = 1$
$\Rightarrow b^2 = 2$
$\Rightarrow\text{b}=\pm\sqrt2$
Hence, a = 1 and $\text{b}=\pm\sqrt2$
View full question & answer→Question 395 Marks
On dividing, $3x^3+ x^2 + 2x + 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
$\therefore$ $3x^3 + x^2 + 2x + 5 = (3x - 5)g(x) + 9x + 10$
$\Rightarrow 3x^3+ x^2 + 2x + 5 - 9x - 10 = (3x - 5)g(x)$
$\Rightarrow 3x^3 + x^2 - 7x - 5 = (3x - 5)g(x)$
$\Rightarrow\text{g}(\text{x})=\frac{3\text{x}^3+\text{x}^2-\text{7x}-5}{\text{3x}-5}$
$\therefore$ $g(x) = x^2 + 2x + 1$ View full question & answer→Question 405 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 $p(x) = x^3 - 6x^2 - x + 30$
View full question & answer→Question 415 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 $3 x^2-4 x+2$ View full question & answer→Question 425 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 435 Marks
Find all the zeros of the polynomial $(2x^4 - 11x^3 + 7x^2 + 13x)$, it being given that two if its zeros are $3+\sqrt2$ and $3-\sqrt2.$
AnswerLet $f(x) = 2x^4 - 11x^3 + 7x^2 + 13x - 7$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- 6x + 7), we get:
$\therefore$ f(x) = 0
$\Rightarrow 2x^4 - 11x^3 + 7x^2 + 13x - 7 = 0$
$\Rightarrow (x^2 - 6x + 7)(2x^2 + x - 1) = 0$
$\Rightarrow\big(\text{x}+3+\sqrt2\big)\big(\text{x}+3-\sqrt2\big)\$\text{2x}-1)(\text{x}+1)=0$
$\Rightarrow\text{x}=-3-\sqrt2$ 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 445 Marks
If 3 and -3 are two zeros of the polynomial $(x^4 + x^3 - 11x^2 - 9x + 18)$, 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 or $x=3$ or $x=-3$ Hence, all the zeros are $1,-2,3$ and -3 View full question & answer→Question 455 Marks
If one zero of the polynomial $p(x) = x^3 - 6x^2 + 11x - 6$ is 3, find the other two zeroes.
AnswerGiven: $p(x) = x^3 - 6x^2 + 11x - 6$ and its factor, x + 3
Let us divided p(x) by (x - 3).
Here, $x^3 - 6x^2 + 11x - 6$
$= (x - 3)(x^2 - 3x + 2)$
$= (x - 3)[x^2 - (2 + 1)x + 2]$
$= (x - 3)(x^2 - 2x - x + 2)$
$= (x - 3)[x(x - 2) - 1(x - 2)]$
$= (x - 3)(x - 1)(x - 2)$
$\therefore$ The other two zeros are 1 and 2.
View full question & answer→Question 465 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 475 Marks
Obtain all other zeros of $(x^4 + 4x^3 - 2x^2 - 20x - 15)$ if two of its zeros are $\sqrt5$ and $-\sqrt5.$
AnswerLet $f(x)=x^4+4 x^3-2 x^2-20 x-15$ Since $\sqrt{5}$ and $-\sqrt{5}$ are the zeros of $f(x)$, it follows that each one of $(x-\sqrt{5})$ and $(x+\sqrt{5})$ is a factor of $f(x)$. Consequently, $(x-\sqrt{5})(x+\sqrt{5})=\left(x^2-5\right)$ is a factor of $f(x)$. On dividing $f(x)$ by $\left(x^2\right.$ - 5), we get:
$\therefore f(x)=0 \Rightarrow x^4+4 x^3-2 x^2-20 x-15 \Rightarrow\left(x^2-5\right)\left(x^2-4 x+3\right) \Rightarrow(x-\sqrt{5})(x+\sqrt{5})(x+1)(x+3)=0 \Rightarrow x=\sqrt{5}$
or $x=-\sqrt{5}$ or $x=-1$ or $x=-3$ Hence, all the zeros are $\sqrt{5},-\sqrt{5},-1$ and -3 View full question & answer→Question 485 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
$\Rightarrow 6x^2 - 3 = 0 \Rightarrow 3(2x^2 - 1) = 0$
$\Rightarrow 2x^2 - 1 = 0$
$\Rightarrow 2x^2 = 1$
$\Rightarrow\text{x}^2=\frac{1}{2}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt2}$
Hence, the zeros of the quadratic polynomial $f(x) = 6x^2 - 3 $are $\frac{1}{\sqrt2},\ -\frac{1}{\sqrt2}.$
View full question & answer→Question 495 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
$\Rightarrow (x^2 + x - 20)(x^2 - 3) = 0$
$\Rightarrow (x^2 + 5x - 4x - 20)(x^2 - 3) = 0$
$\Rightarrow [x(x + 5) - 4(x + 5)](x^2 - 3) = 0$
$\Rightarrow(\text{x}-4)(\text{x}+5) \big(\text{x}-\sqrt3\big)\big(\text{x}+\sqrt3\big)=0$
⇒ 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 505 Marks
Use remainder theorem to find the value of k, it being given that when $x^3 + 2x^2 + kx + 3$ is divided by (x - 3), then the remainder is 21.
AnswerLet $p(x) = x^3 + 2x^2 + kx + 3$
$Now, p(3) = (3)^3 + 2(3)^2 + 3k + 3$
$= 27 + 18 + 3k + 3$
$= 48 + 3k$
It is given that the remainder is 21
$\therefore$ $3k + 48 = 21$
$\Rightarrow 3k = -27$
$\Rightarrow k = -9$
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