Question 12 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=x^3-8, g(x)=x-2$
Answer$f(x)=\left(x^3-8\right)$
By the Factor Theorem, $(x-2)$ will be a factor of $f(x)$ if $f(2)=0$.
Here, $f(2)=(2)^3-8$
$=8-8=0$
$\therefore(\mathrm{x}-2)$ is a factor of $\left(\mathrm{x}^3-8\right)$.
View full question & answer→Question 22 Marks
Give an example of a monomial of degree $0.$
AnswerA polynomial having one term is called a monomial. Since the degree of required monomial is $0$, so the highest power of $x$ in the monomial should be $0$. An example of a monomial of degree $0$ is $5.$
View full question & answer→Question 32 Marks
Give an example of a binomial of degree $8.$
AnswerA polynomial having two term is called a binomial. Since the degree of required binomial is $8$ , so the highest power of $x$ in the binomial should be $8 .$
An example of a binomial of degree 8 is $2 x^8-3 x$
View full question & answer→Question 42 Marks
Find the value of a for which $(x+1)$ is a factor of $\left(a x^3+x^2-2 x+4 a-9\right)$.
AnswerLet $p(x)=a x^3+x^2-2 x+4 a-9$
It is given that $(x+1)$ is a factor of $p(x)$.
$\Rightarrow \mathrm{p}(-1)=0$
$\Rightarrow \mathrm{a}(-1)^3+(-1)^2-2(-1)+4 a-9=0$
$\Rightarrow-\mathrm{a}+1+2+4 a-9=0$
$\Rightarrow 3 \mathrm{a}-6=0$
$\Rightarrow 3 \mathrm{a}=6$
$\Rightarrow \mathrm{a}=2$
View full question & answer→Question 52 Marks
Verify that:
$1$ and $2$ are the zeros of the polynomial $p(x)=x^2-3 x+2$.
Answer$p(x)=x^2-3 x+2=(x-1)(x-2)$
$\Rightarrow p(1)=(1-1) \times(1-2)$
$=0 \times(-1)$
$=0$
Also,
$p(2)=(2-1)(2-2)$
$=(-1) \times 0$
$=0$
Hence, $1$ and $2$ are the zeroes of the given polynomial.
View full question & answer→Question 62 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=2 x^4+9 x^3+6 x^2-11 x-6, g(x)=x-1$
Answer$f(x)=\left(2 x^4+9 x^3+6 x^2-11 x-6\right)$
By the Factor Theorem, $(x-1)$ will be a factor of $f(x)$ if $f(1)=0$.
Here, $f(1)=2 \times 1^4+9 \times 1^3+6 \times 1^2-11 \times 1-6$
$=2+9+6-11-6$
$=17-17=0$
$\therefore(x-1)$ is factor of $\left(2 x^4+9 x^3+6 x^2-11 x-6\right)$.
View full question & answer→Question 72 Marks
Give an example of a trinomial of degree $4.$
AnswerA polynomial having three term is called a trinomial. Since the degree of required binomial is $4 $, so the highest power of $x$ in the trinomial should be $4 .$
An example of a trinomial of degree 4 is $2 x^4-3 x+5$
View full question & answer→Question 82 Marks
Find the value of a for which $(x-4)$ is a factor of $\left(2 x^3-3 x^2-18 x+a\right)$.
Answer$f(x)=\left(2 x^3-3 x^2-18 x+a\right)$
$x-4=0$
$\Rightarrow x=4$
$\therefore f(4)=2(4)^3-3(4)^2-18 \times 4+a$
$=128-48-72+a$
$=128-120+a$
$=8+a$
Given that $(x-4)$ is a factor of $f(x)$.
By the Factor Theorem, $(x-a)$ will be a factor of $f(x)$ if $f(a)=0$ and therefore $f(4)=0$.
$\Rightarrow f(4)=8+a=0$
$\Rightarrow a=-8$
View full question & answer→Question 92 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=2 x^3+7 x^2-24 x-45, g(x)=x-3$
Answer$f(x)=\left(2 x^3+7 x^2-24 x-45\right)$
By the Factor Theorem, $(x-3)$ will be a factor of $f(x)$ if $f(3)=0$.
Here, $f(3)=2 \times 3^3+7 \times 3^2-24 \times 3-45$
$=54+63-72-45$
$=117-117=0$
$\therefore(x-3)$ is a factor of $\left(2 x^3+7 x^2-24 x-45\right)$.
View full question & answer→Question 102 Marks
Find the value of a for which $(x+2 a)$ is a factor of $\left(x^5-4 a^2 x^3+2 x+2 a+3\right)$
AnswerLet $p(x)=x^5-4 a^2 x^3+2 x+2 a+3$
It is given that $(x+2 a)$ is a factor of $p(x)$.
$\Rightarrow p(-2 a)=0$
$\Rightarrow(-2 a)^5-4 a^2(-2 a)^3+2(-2 a)+2 a+3=0$
$\Rightarrow-32 a^5-4 a^2\left(-8 a^3\right)-4 a+2 a+3=0$
$\Rightarrow-32 a^5+32 a^5-2 a+3=0$
$\Rightarrow 2 a=3$
$\Rightarrow \mathrm{a}=\frac{3}{2}$
View full question & answer→Question 112 Marks
Find the value of a for which the polynomial $\left(x^4-x^3-11 x^2-x+a\right)$ is divisible by $(x+3)$.
AnswerLet $\mathrm{p}(\mathrm{x})=\mathrm{x}^4-\mathrm{x}^3-11 \mathrm{x}^2-\mathrm{x}+\mathrm{a}$
It is given that $p(x)$ is divisible by $(x+3)$.
$\Rightarrow(x+3) \text { is a factor of } p(x)$
$\Rightarrow p(-3)=0$
$\Rightarrow(-3)^4-(-3)^3-11(-3)^2-(-3)+a=0$
$\Rightarrow 81+27-99+3+a=0$
$\Rightarrow 12+a=0$
$\Rightarrow a=-12$
View full question & answer→Question 122 Marks
Give an example of a monomial of degree $5.$
AnswerA polynomial having one term is called a monomial. Since the degree of required monomial is $5$ , so the highest power of $x$ in the monomial should be $5 .$
An example of a monomial of degree 5 is $2 x^5$.
View full question & answer→Question 132 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when$\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$ $\text{g}(\text{x})=\text{x}+\sqrt2$
Answer$\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$By the factor theorem, $(x - a)$ will be factor of $p(x)$ if $p(a) = 0.$
Here, $\text{f}(-\sqrt{2})=2\sqrt2(-\sqrt2)^2+5(-\sqrt2)+\sqrt2$
$=2\sqrt2\times2-5\sqrt2+\sqrt2$
$=4\sqrt2-5\sqrt2+\sqrt2$
$=5\sqrt2-5\sqrt2=0.$
$\therefore(\text{x}+\sqrt2)$ is a factor of $2\sqrt2\text{x}^2+5\text{x}+\sqrt2.$
View full question & answer→Question 142 Marks
Find the value of $k$ for which $(x-1)$ is a factor of $\left(2 x^3+9 x^2+x+k\right)$.
Answer$f(x)=\left(2 x^3+9 x^2+x+k\right) x-1=0 $
$\Rightarrow x=1 \therefore f(1)=2 \times 1^3+9 \times 1^2+1+k=2+9+1+k=12+k$
Given that $(x-1)$ is a factor of $f(x)$. By the Factor Theorem, $(x-a)$
will be a factor of $f(x)$ if $f(a)=0$ and therefore $f(1)=0 . $
$\Rightarrow f(1)=12+k$
$=0 $
$\Rightarrow k=-12$.
View full question & answer→Question 152 Marks
Rewrite the following polynomial in standard form.
$x-2 x^2+8+5 x^3$
Answer$8+x-2 x^2+5 x^3$ is a polynomial in standard form as the powers of $x$ are in ascending order.
View full question & answer→Question 162 Marks
Verify that:
$0$ and $3$ are the zeros of the polynomial, $r(x)=x^2-3 x$.
Answer$r(x)=x^2-3 x$
$r(0)=0^2-3 x 0$
Also,
$r(3)=3^2-3 \times 3$
$=9-9$
$=0$
Hence, $0$ and $3$ are the zeroes of the given polynomial.
View full question & answer→Question 172 Marks
Using factor theorem, show that $g(x)$ is a factor of p(x), when $\text{p}(\text{x})=7\text{x}^2-4\sqrt2\text{x}-6,\ $ $\text{g}(\text{x})=\text{x}-\sqrt2$
View full question & answer→Question 182 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=2 x^3+9 x^2-11 x-30, g(x)=x+5$
Answer$f(x)=2 x^3+9 x^2-11 x-30$
By the Factor Theorem, $(x+5)$ will be a factor of $f(x)$ if $f(-5)=0$.
Here, $f(-5)=2(-5)^3+9(-5)^2-11(-5)-30$
$=-250+225+55-30$
$=-280+280=0$
$\therefore(x+5)$ is a factor of $\left(2 x^3+9 x^2-11 x-30\right)$.
View full question & answer→Question 192 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=69+11 x-x^2+x^3, g(x)=x+3$
AnswerBy the factor theorem, $g(x)=x+3$ will be a factor of $p(x)$ if $p(-3)=0$.
Now, $p(x)=69+11 x-x^2+x^3$
$\Rightarrow \mathrm{p}(-3)=69+11(-3)-(-3)^2+(-3)^3$
$=69-33-9-27$
$=0$
Hence, $g(x)=x+3$ is a factor of the given polynomial $p(x)$.
View full question & answer→Question 202 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=x^4-x^2-12, g(x)=x+2$
Answer$f(x)=\left(x^4-x^2-12\right)$
By the Factor Theorem, $(x+2)$ will be a factor of $f(x)$ if $f(-2)=0$.
Here, $f(-2)=(-2)^4-(-2)^2-12$
$=16-4-12$
$=16-16=0$
$\therefore(\mathrm{x}+2)$ is a factor of $\left(\mathrm{x}^4-\mathrm{x}^2-12\right)$.
View full question & answer→Question 212 Marks
Verify that: $2$ and $-3$ are the zeros of the polynomial $q(x)=x^2+x-6$.
Answer$q(x)=x^2+x-6 \Rightarrow q(2)=2^2+2-6=4-4=0$ Also, $q(-3)=(-3)^2+(-3)-6=9-9=0$ Hence, $2$ and $-3$ are the zeroes of the given polynomial.
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