4 Marks Questions

Question 14 Marks

Without actual division, prove that $2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2.$

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, $f(2)=2(2)^4-5(2)^3+2(2)^2-2+2=32-40+8=0$ 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 $p(x) = 2x^4 + x^3 - 8x^2 - x + 6, g(x) = 2x - 3$

Answer

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

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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 $
$\Rightarrow b=-14$
$\therefore a=23 \text { and } b=-14$

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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, $x^2+2 x-3$
$= x^2 + 3x - x - 3$
$= x (x + 3) - 1 (x + 3)$
$= (x + 3) (x - 1)$
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, $x^2 + 2x - 3$ divides $f(x)$ exactly.

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Question 54 Marks

By actual division, find the quotient and the remainder when $\left(x^4+1\right)$ 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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