4 Mark Question

Question 14 Marks

VWithout actual division, prove that $2 x^4-5 x^3+2 x^2-x+2$ is divisible by $x^2-3 x+2$.

Answer

Let $f(x)=2 x^4-5 x^3+2 x^2-x+2$
$g(x)=x^2-3 x+2$
$=x^2-2 x-x+2$
$=x(x-2)-1(x-2)$
$=(x-2)(x-1)$
Clearly, $(x-2)$ and $(x-1)$ are factors of $g(x)$.
In order to prove that $f(x)$ is exactly divisible by $g(x)$, it is sufficient to prove that $f(x)$ is exactly divisible by $(x-2)$ and $(x-1)$.
Thus, we will show that $(x-2)$ and $(x-1)$ are factors of $f(x)$.
Now,
$f(2)=2(2)^4-5(2)^3+2(2)^2-2+2$
$=32-40+8=0 \text { and }$
$f(1)=2(1)^4-5(1)^3+2(1)^2-1+2$
$=2-5+2-1+2=0$
Therefore, $(x-2)$ and $(x-1)$ are factors of $f(x)$.
$\Rightarrow g(x)=(x-2)(x-1)$ is a factor of $f(x)$.
Hence, $f(x)$ is exactly divisible by $g(x)$.

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Question 24 Marks

Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=2 x^4+x^3-8 x^2-x+6, g(x)=2 x-3$

Answer

$f(x)=\left(2 x^4+x^3-8 x^2-x+6\right)$ By the Factor Theorem, $(x-a)$ will be a factor of $f(x)$ if $f(a)=0$.
Here, $2 x-3=0 x=\frac{3}{2}$
$f\left(\frac{3}{2}\right)=2\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^3-8\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)+6 \\
=2 \times \frac{81}{16}+\frac{27}{8}-8 \times \frac{9}{4}-\frac{3}{2}+6 \\
=\frac{81}{8}+\frac{27}{8}-18-\frac{3}{2}+6 \\
=\frac{81+27-144-12+48}{8} \\
=\frac{156-156}{8}=0$
$\therefore(2 x -3)$ is a factor of $\left(2 x ^4+ x ^3-8 x ^2- x +6\right)$.

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Question 34 Marks

Find the values of $a$ and $b$ so that the polynomial $\left(x^3-10 x^2+a x+b\right)$ is exactly divisible by $(x-1)$ as well as $(x-2)$.

Answer

Let $f(x)=\left(x^3-10 x^2+a x+b\right)$,
then by factor theorem $(x-1)$ and $(x-2)$ will be factors of $f(x)$ if $f(1)=0$ and $f(2)=0$.
$\therefore f(1)=1^3-10 \times 1^2+a \times 1+b=0$
$\Rightarrow 1-10+a+b=0$
$\Rightarrow a+b=9 \ldots \text { (i) And f( } 2)=2^3-10 \times 2^2+a \times 2+b=0$
$\Rightarrow 8-40+2 a+b=0$
$\Rightarrow 2 a+b=32 \ldots \text { (ii) Subtracting (i) from (ii), }$
we get $a=23$ Substituting the value of $a=23$ in (i),
we get $23+b=9$
$\Rightarrow b=9-23$
$\Rightarrow b=-14$
$\therefore a=23$ and $b=-14$

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Question 44 Marks

Without actual division, show that $\left(x^3-3 x^2-13 x+15\right)$ is exactly divisible by $\left(x^2+2 x-3\right)$

Answer

Let $f(x)=x^3-3 x^2-13 x+15$
Now, $x^2+2 x-3$
$=x^2+3 x-x-3$
$=x(x+3)-1(x+3)$
$=(x+3)(x-1)$
Thus, $f(x)$ will be exactly divisible by
$x^2+2 x-3$
$=(x+3)(x-1)$ if $(x+3)$ and $(x-1)$ are both factors of $f(x)$,
so by factor theorem, we should have $f(-3)=0$ and $f(1)=0$.
Now, $f(-3)=(-3)^3-3(-3)^2-13(-3)+15$
$=-27-3 \times 9+39+15$
$=-27-27+39+15=-54+54$
$=0$ And,
$f (1)=1^3-3 \times 1^2-13 \times 1+15$
$=1-3-13+15=16-16=0$
$\therefore f (-3)=0$ and $f (1)=0$
So, $x^2+2 x-3$ divides $f(x)$ exactly.

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Question 54 Marks

By actual division, find the quotient and the remainder when $(x^4 + 1)$ is divided by $(x - 1).$
Verify that remainder =$ f(1).$

Answer

Let $f(x)= x^4 + 1$ and $g(x) = x - 1.$

Quotient $= x^3 + x^2 + x + 1$
Remainder =$ 2$
Verification:
Putting $x = 1$ in $f(x),$
we get
$f(1) = 1^4 + 1 = 1 +1 = 2$ = Remainder,
when $f(x) = x^4 + 1$ is divided by $g(x) = x - 1.$

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