Question 12 Marks
What must be subtracted from $x^3-6 x^2-15 x+80$ so that the result is exactly divisible by $x^2+x-12 ?$
Answer
$\therefore 4 x-4$ should subtracted from $x^3-6 x^2-15 x+80$
So that the result is exactly divisible by $x^2+x-12$. View full question & answer→Question 22 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$f(x)=x^2, x=0$
Answer$f(x)=x^2, x=0$
we know that, $f(x)=x^2$
Given that value of $x$ is ' 0 '
Substitute the value of $x$ in $f(x)$
$f(0)=0^2$
$=0$
Since, the result is zero, $x=0$ is the root of $x^2$
View full question & answer→Question 32 Marks
Find the remainder when $x^3+3 x^3+3 x+1$ is divided by:
x
AnswerHere,
$f(x)=x^3+3 x^2+3 x+1$
By remainder theorem
$x=0$
substitute the value of $x$ in $f(x)$
$f(0)=0^3+3(0)^2+3(0)+1$
$=0+0+0+1$
$=1$
View full question & answer→Question 42 Marks
In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
$f(x) = 3x^4 + 17x^3 + 9x^2 - 7x - 10; g(x) = x + 5$
AnswerLet $g(x)$
$=0 \Rightarrow x+5$
$=0 \Rightarrow x=-5$
Now, $f(-5)=3(-5)^4+17(-5)^3+9(-5)^2-7(-5)-10$
$=3(625)+17(-125)+9(25)+35-10$
$=1875-2125+225+35-10$
$\because f(-5)=0$, by factor theorem $x+5$ is a factor of $f(x)$.
View full question & answer→Question 52 Marks
If $x + 1$ is a factor of $x^3 + a$, then write the value of a.
AnswerAs $(x + 1)$ is a factor of polynomial $f(x) = x^3 + a.$
i. e. $f(-1) = 0$
$(-1)^3+ a = 0$
$\Rightarrow a = 1$
Thus, the value of a = 1.
View full question & answer→Question 62 Marks
If $\text{x}=\frac{1}{2}$ is a zero of the polynomial $f(x) = 8x^3 + ax^2 - 4x + 2$, find the value of a.
AnswerSince $\text{x}=\frac{1}{2}$ is a zero of polynomial f(x).
Therefore $\text{f}\Big(\frac{1}{2}\Big)=0$$\Rightarrow\ 8\Big(\frac{1}{2}\Big)^3+\text{a}\Big(\frac{1}{2}\Big)^2-4\Big(\frac{1}{2}\Big)+2=0$
$\Rightarrow\ 1+\frac{\text{a}}{4}-2+2=0$
$\Rightarrow\ \text{a}=-4$
The value of a is -4.
View full question & answer→Question 72 Marks
If f(x) =$2x^3 - 13x^2 + 17x + 12$, Find:
$f(-3)$
AnswerThe given polynomial is f(x) = $2x^3 - 13x^2 + 17x + 12$
$f(-3)$
we need to substitute the $'(-3)'$ in $f(x)$
$f(-3) = 2(-3)^3 - 13(-3)^2 + 17(-3) + 12$
$= (2 \times (- 27)) - (13 \times 9) - ( 17 \times 3) + 12$
$= -54 - 117 - 51 + 12$
$= -210$
therefore $f(-3) = -210$
View full question & answer→Question 82 Marks
In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
$f(x) = x^3 - 6x^2 - 19x + 84, g(x) = x - 7$
AnswerLet $g(x) = 0$
$\Rightarrow x - 7 = 0$
$\Rightarrow x = 7$
$f(7) = 7^3 - 6(7)^2- 19(7) + 84$
$= 343 - 6(49) - 19(7) + 84$
$= 343 - 294 - 133 + 84$
$= 427 - 427 = 0$
$\because$ $f(7) = 0$, by factor theorem $x - 7$ is a factor of $f(x).$
View full question & answer→Question 92 Marks
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by:
$x + 1$
AnswerHere,$ f(x) = x^3 + 3x^2 + 3x + 1$
By remainder theorem
$x + 1 = 0$
$\Rightarrow x = -1$
substitute the value of x in f(x)
$f(-1) = (−1)^3 + 3(−1)^2 + 3(−1) + 1$
$= -1 + 3 - 3 + 1$
$= 0$
View full question & answer→Question 102 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x)=x^5+3 x^4-x^3-3 x^2+5 x+15, g(x)=x+3$
AnswerLet $g(x)=0$
$\Rightarrow x+3=0$
$\Rightarrow x=-3$
$f(-3)=(-3)^5-3(-3)^4-(-3)^3-3(-3)^2+5(-3)+15$
$=-243+243+27-27-15+15=0$
$\because f(-3)=0, \text { by factor theorem } x+3 \text { is a factor of } f(x) \text {. }$
View full question & answer→Question 112 Marks
If $f(x)=x^4-2 x^3+3 x^2-a x-b$ when divided by $x-1$, the remainder is 6 , then find the value of $a+b$.
AnswerWhen polynomial $f(x)=x^4-2 x^3+3 x^2-a x-b$ divided by $(x-1)$
The remainder is 6 .
i. e. $f(1)=6$
$(1)^4-2(1)^3+3(1)^2-a(1)-b=6$
$1-2+3-a-b=6$
$2-(a+b)=6$
$(a+b)=-4$
Thus, the value of $a+b=-4$.
View full question & answer→Question 122 Marks
In the following two polynomials, find the value of a, if $x + a$ is a factor.
$x^3 + ax^2 - 2x + a + 4$
Answer$x^3 + ax^2 - 2x + a + 4$
Let, $x + a = 0 \Rightarrow x = -a$
$\because$ $(x + a)$ is a factor of $f(x) = x^3 + ax^2 - 2x + a + 4$
$\therefore$ $p(-a) = 0$
$p(-a) \Rightarrow (-a)^3 + (-a)^2 - 2(-a) + a + 4$
$= 0 \Rightarrow -a^3 + a^3 + 2a + a + 4$
$= 0 \Rightarrow 3a + 4$
$= 0 \Rightarrow 3a$
$= -4$ $\Rightarrow\ \text{a}=\frac{-4}{3}$
View full question & answer→Question 132 Marks
If $f(x)=2 x^3-13 x^2+17 x+12$, Find:
f(2)
AnswerThe given polynomial is $f(x)=2 x^3-13 x^2+17 x+12$ f(2)
we need to substitute the ' 2 ' in $f(x)$
$f(2)=2(2)^3-13(2)^2+17(2)+12$
$=(2 \times 8)-(13 \times 4)+(17 \times 2)+12$
$=16-52+34+12$
$=10$
therefore $f(2)=10$
View full question & answer→Question 142 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:$\text{f(x)}=\text{l(x)}+\text{m},\text{x}=-\frac{\text{m}}{\text{l}}$
Answer$\text{f(x)}=\text{l(x)}+\text{m},\text{x}=-\frac{\text{m}}{\text{l}}$We know that,
$\text{f(x)}=\text{l(x)}+\text{m}$
Given, that $\text{x}=-\frac{\text{m}}{\text{l}}$
Substitute the value of x in f(x)
$\text{f}\Big(-\frac{\text{m}}{\text{l}}\Big)=\text{I}\Big(-\frac{\text{m}}{\text{l}}\Big)+\text{m}$
$=-\text{m}+\text{m}$
$=0$
Since, the result is 0, $\text{x}=-\frac{\text{m}}{\text{l}}$ is the root of lx + m
View full question & answer→Question 152 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:$\text{f(x)}=3\text{x}+1,\text{x}=-\frac{1}{3}$
Answer$\text{f(x)}=3\text{x}+1,\text{x}=-\frac{1}{3}$We know that,
$\text{f(x)}=3\text{x}+1$
Substitite $\text{x}=-\frac{1}{3}$ in f(x)
$\text{f}\Big(-\frac{1}{3}\Big)=3\Big(-\frac{1}{3}\Big)+1$
$= -1 + 1$
$=0$
Since, the result is 0 $\text{x}=-\frac{1}{3}$ is the root of 3x + 1
View full question & answer→Question 162 Marks
Give one example each of a binomial of degree $35$, and of a monomial of degree $100$
AnswerGiven, to write the examples for binomial and monomial with the given degrees.
Example of a binomial with degree $35-7 x^{35}-5$
Example of a monomial with degree $100-2 t^{100}$
View full question & answer→Question 172 Marks
In the following two polynomials, find the value of $a$, if $x - a$ is factor:
$x^5-a^2 x^3+2 x+a+1$
Answer$x^5-a^2 x^3+2 x+a+1$
Let, $x-a=0 \Rightarrow x=a$
$\because(x-a)$ is a factor of $f(x)=x^5-a^2 x^3+2 x+a+1$
$\therefore f(a)=0$
$f(a) \Rightarrow a^5-a^2 \times a^3+2 a+a+1$
$=0 \Rightarrow a^5-a^5+3 a+1 \Rightarrow 3 a=-1$
$\Rightarrow a=-\frac{1}{3}$
View full question & answer→Question 182 Marks
In the following two polynomials, find the value of $a$, if $x-a$ is factor:
$x^6-a x^5+x^4-a x^3+3 x-a+2$
Answer$x^6-a x^5+x^4-a x^3+3 x-a+2$
Let, $x - a =0 \Rightarrow x = a$
$\because x - a$ is a factor of the polynomial $f ( x )$
$\therefore f(a)=0$
$f(a)=0$
$\Rightarrow a^6-a \times a^5+a^4-a \times a^3+3 a-a+2=0$
$\Rightarrow a^6-a^6+a^4-a^4+3 a-a+2=0$
$\Rightarrow 2 a+2=0$
$\Rightarrow 2 a=-2$
$\Rightarrow a=-1$
View full question & answer→Question 192 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial
$f(x)$ or, not: $f(x)=x^3-6 x^2+11 x-6 ; g(x)=x-3$
AnswerLet $g ( x )=0$
$\Rightarrow x-3=0$
$\Rightarrow x=3$
$f(3)=3^3-6(3)^2+11(3)-6$
$=27-6(9)+33-6$
$=60-60$
$=0$
$\because f(3)=0$, by factor theorem $x-3$ is a factor of $f(x)$.
View full question & answer→Question 202 Marks
In the following two polynomials, find the value of $a$, if $x+a$ is a factor.
$x^4-a^2 x^2+3 x-a$
Answer$x^4-a^2 x^2+3 x-a$
Let,
$x+a=0$
$\Rightarrow x=-a$
$\because(x+a) \text { is a factor of } f(x)=x^4-a^2 x^2+3 x-a$
$\therefore f(-a)=0$
$f(-a) \Rightarrow(-a)^4-a^2(-a)^2+3(-a)-a=0$
$\Rightarrow a^4+a^4-3 a-a=0$
$\Rightarrow-4 a=0$
$\Rightarrow a=0$
View full question & answer→Question 212 Marks
Find the remainder when $x^3+3 x^3+3 x+1$ is divided by: $x-\frac{1}{2}$
AnswerHere, $f(x)=x^3+3 x^2+3 x+1$ By remainder theorem $x-\frac{1}{2}$
$\Rightarrow x=\frac{1}{2}$
substitute the value of $x$ in $f(x) f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^3+3\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)+1$
$=\left(\frac{1}{2}\right)^3+3\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)+1$
$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$
$=\frac{1+6+12+8}{8}$
$=\frac{27}{8}$
View full question & answer→Question 222 Marks
If $f(x)=2 x^3-13 x^2+17 x+12$,
Find: $f (0)$
AnswerThe given polynomial is $f(x)=2 x^3-13 x^2+17 x+12$
$f(0)$
we need to substitute the ' $(0)^{\prime}$ in $f(x)$
$f(0)=2(0)^3-13(0)^2+17(0)+12$
$=(2 \times 0)-(13 \times 0)+(17 \times 0)+12$
$=0-0+0+12$
$=12$
therefore $f(0)=12$
View full question & answer→