[5 marks sum]

Question 15 Marks

Given $f(x)=a x^2+b x+2$ and $g(x)=b x^2+a x+1$. If $x-2$ is a factor of $f(x)$ but leaves the remainder -15 when it divides $g(x)$, find the values of $a$ and $b$. With these values of $a$ and $b$, factorise the expression. $f(x)+g(x)+4 x^2+7 x$.

Answer

$ f(x)=a x^2+b x+2 $
$ g(x)=b x^2+a x+1 $
$ x-2 \text { is } a \text { factor of } f(x) $
$ \text { Let } x-2=0 $
$ \Rightarrow x=2 $
$ \therefore f(2)=a(2)^2+b \times 2+2=4 a+2 b+2 $
$ \therefore 4 a+2 b+2=0 \quad \ldots(\because x-2 s \text { its factor) } $
$ \Rightarrow 2 a+b+1=0 \quad \ldots(\text { (i) (Dividing by } 2) $
$ \text { Dividing } g(x) \text { by } x-2, \text { remainder }=-15 $
$ \text { Let } x-2=0 $
$ \Rightarrow x=2 $
$ \therefore g(2)=b(2)^2+a x 2+1 $
$ =4 b+2 a+1 $
$ \because \text { Remainder is }-15 $
$ \therefore 4 b+2 a+1=-15 $
$ \Rightarrow 4 b+2 a+1+15=0 $
$ \Rightarrow 4 b+2 a+16=0 $
$ \Rightarrow 2 b+a+8=0 \quad \ldots \text { (Dividing by } 2 \text { ) } $
$ \Rightarrow a+2 b+8=0 $
$ \text { Multiplying (i) by } 2 \text { and (ii) by } 1 $
$ 4 a+2 b+2=0 $
$ a +2 b+8=0$

Substituting the value of $a$ in (i)
$2 \times 2+b+1=0$
$\Rightarrow 4+b+1=0$
$\Rightarrow b+5=0$
$\Rightarrow b=-5$
Hence $a=2, b=-5$
Now $f ( x )+ g ( x )=4 x 2+7 x$
$=2 x^2-5 x+2+\left(-5 x^2+2 x+1\right)+4 x^2+7 x$
$=2 x^2-5 x+2-5 x^2+2 x+1+4 x^2+7 x$
$=6 x^2-5 x^2-5 x+2 x+7 x+2+1$
$=x^2+2 x+3$
$=x^2+x+3 x+3$
$=x(x+1)+3(x+1)$
$=(x+1)(x+3) .$

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Question 25 Marks

If $ax^3 + 3x^2 + bx – 3$ has a factor $(2x + 3)$ and leaves remainder $– 3$ when divided by $(x + 2)$, find the values of a and b. With these values of a and b, factorise the given expression.

Answer

Let,
$p(x)=a x^3+3 x^2+b x-3$
$g(x)=2 x+3=0 \Rightarrow x=-\frac{3}{2}$
$f(x)=x+2=0 \Rightarrow x=-2$
Given, $g(x)$ is a factor of $f(x)$
$\therefore$ By factor theorem,
$p\left(-\frac{3}{2}\right)=0$
$\Rightarrow a\left(-\frac{3}{2}\right)^3+3\left(-\frac{3}{2}\right)^2+b\left(-\frac{3}{2}\right)-3=0$
$\Rightarrow a\left(-\frac{27}{8}\right)+3\left(\frac{9}{4}\right)+b\left(-\frac{3}{2}\right)-3=0$
$\Rightarrow-\frac{27 a}{8}+\frac{27}{4}-\frac{3 b}{2}-3=0$
$\Rightarrow-\frac{27 a+54-12 b}{8}=3$
$\Rightarrow-27 a+54-12 b=24$
$\Rightarrow-3(9 a+4 b)=24-54=-30$
$\Rightarrow 9 a+4 b=10.....(1)$
Also, $p(x)$ when divided by $f(x)$ leaves a remainder -3
$\therefore$ By remainder theorem,
$p(-2)=-3$
$\Rightarrow a(-2)^3+3(-2)^2+b(-2)-3=-3$
$\Rightarrow-8 a+12-2 b=0$
$\Rightarrow 8 a+2 b=12$
$\Rightarrow 4 a+b=6......(2)$
Solving (i) and (ii), we get $a=2$ and $b=-2$
$\text { Hence, } p(x)=2 x^3+3 x^2-2 x-3$
$=x^2(2 x+3)-(2 x+3)$
$=(2 x+3)\left(x^2-1\right)$
$=(2 x+3)(x+1)(x-1)$

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Question 35 Marks

If $(x-2)$ is a factor of the expression $2 x^3+a x^2+b x-14$ and when the expression is divided by ( $x-3$ ), it leaves a remainder 52 , find the values of $a$ and $b$.

Answer

$
\begin{aligned}
& f(x)=2 x^3+a x^2+b x-14 \\
& \therefore(x-2) \text { is factor of } f(x) \\
& f(2)=0 \\
& 2(2)^3+a(2)^2+b(2)-0 \\
& 16+4 a+2 b-14=0 \\
& \Rightarrow 4 a+2 b=-2 \\
& 2 a+b=-1
\end{aligned}
$
Also, $(x-3)$ it leaves remainder $=52$
$
\begin{aligned}
& \therefore f(3)=52 \\
& 2(3)^3+a(3)^2+b(3)-14=52 \\
& \Rightarrow 54+9 a+3 b-14=52 \\
& \Rightarrow 9 a+3 b=52-40 \\
& 9 a+3 b=12 \\
& 3 a+b=4
\end{aligned}
$
From (i) and (ii)
$
\begin{aligned}
& 2 a+b=-1 \\
& 3 a+b=4
\end{aligned}
$
Subtracting $-a \quad=-5$
$
\begin{aligned}
& \therefore a=5 \text { put in (i) } \\
& \therefore 2(5)+b=-1 \\
& \Rightarrow b=-1-10 \\
& \Rightarrow b=-11 \\
& a=5, b=-11 \\
& \Rightarrow \frac{-27 a}{8}+\frac{27}{4}-\frac{3 b}{2}=0 \\
& \Rightarrow-27 a+54-12 b-24=0 \quad \ldots \text {(Multiplying by } 8) \\
& \Rightarrow-27 a-12 b+30=0 \\
& \Rightarrow-27 a-12 b=-30 \\
& \Rightarrow 9 a+4 b=10 \quad \ldots[\text { [Dividing by (-3)] } \\
& 9 a+4 b=10
\end{aligned}
$
Again let $x+2=0$ then $x=-2$
Substituting the value of $x$ in $f(x)$
$
\begin{aligned}
& f(x)=a x^3+3 x^2+b x-3 \\
& f(-2)=a(-2)^3+3(-2)^2+b(-2)-3 \\
& =-8 a+12-2 b-3 \\
& =-8 a-2 b+9 \\
& \because \text { Remainder }=-3 \\
& \therefore-8 a-2 b+9=-3 \\
& \Rightarrow-8 a-2 b=-3-9 \\
& \Rightarrow-8 a-2 b=-12 \quad \ldots \text { (Dividing by } 2) \\
& \Rightarrow 4 a+b=6
\end{aligned}
$
Multiplying (ii) by 4
$
\begin{array}{r}
16 a+4 b=24 \\
9 a+4 b=10
\end{array}
$

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Question 45 Marks

If $(x + 2)$ and $(x – 3)$ are factors of $x^3 + ax + b$, find the values of a and b. With these values of a and b, factorise the given expression.

Answer

Let $x+2=0$, then $x=-2$
Substituting the value of $x$ in $f(x)$,
$f(x)=x^3+a x+b$
$f(-2)=(-2)^2+a(-2)+b$
$=-8-2 a+b$
$\because x +2$ is a factor
$\therefore$ Remainder is zero
$\therefore-8-2 a+b=0$
$\Rightarrow-2 a+b=8$
$\therefore 2 a-b=-8$
Again let $x-3=0$, then $x=3$
Substituting the value of $x$ in $f(x)$,
$f(x)=x^3+a x+b$
$f(3)=(3)^3+a(3)+b$
$=27+3 a+b$
$\because x -3$ is a factor
$\therefore$ Remainder $=0$
$\Rightarrow 27+3 a+b=0$
$\Rightarrow 3 a+b=-27$
Adding (i) and (ii)
$5 a=-35$
$\Rightarrow a=\frac{-35}{5}=-7$
Substituting value of $a$ in (i)
$2(-7)-b=-8$
$\Rightarrow-14-b=-8$
$-b=-8+14$
$\Rightarrow-b=6$
$\therefore b=-6$
Hence $a=-7, b=-6$
$(x+2)$ and $(x-3)$
are the factors of
$x ^3+ ax + b$
$\Rightarrow x ^3-7 x -6$
Now dividing $x^3-7 x-6$ by $(x+2)$

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Question 55 Marks

f $2 x^3+a x^2-11 x+b$ leaves remainder $0$ and $42 $when divided by $(x-2)$ and $(x-3)$ respectively, find the values of a and $b$. With these values of $a$ and $b$, factorize the given expression.

Answer

$f(x)=2 x^3+a x^2-11 x+b$
Let $x-2=0$, then $x=2$,
Substituting the vaue of $x$ in $f(x)$
$f(2)=2(2)^3+a(2)^2-11(2)+b$
$\because$ Remainder $=0$,
$ \therefore 4 a+b-6=0$
$ \Rightarrow 4 a+b=6$
Again let $x-3=0$,
then $x=3$
Substituting the value of $x$ is $f(x)$
$f(3)=2(3) 3+a(3) 2-11 \times 3+b$
$=2 \times 27+9 a-33+b$
$=54+9 a-33+b$
$\Rightarrow 9 a+b+21$
$\because \text { Remainder }=42$
$\therefore 9 a+b+21=42$
$\Rightarrow 9 a+b=42-21$
$\Rightarrow 9 a+b=21$
Subtracting (i) from (ii)
$5 a=15$
$\Rightarrow \frac{15}{5}=3$
Substituting the value of $a$ is (i)
$4(3)+b=6$
$\Rightarrow 12+b=6$
$\Rightarrow b=6-12$
$\Rightarrow b=-6$
$\therefore f(x)$ will be $2 x^3+3 x^2-11 x-6$
$\because x-2$ is a factor (as remainder $=0$ )
$\therefore$ Dividing $f ( x )$ by $x -2$, we get

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Question 65 Marks

If $(x+3)$ and $(x-4)$ are factors of $x^3+a x^2-b x+24$, find the values of $a$ and $b$ : With these values of $a$ and $b$, factorise the given expression.

Answer

$f(x)=x^3+a x^2-b x+24$
Let $x+3=0$, then $x=-3$
Substituting the value of $x$ in $f(x)$
$f(-3)=(-3) 3+a(-3) 2-b(-3)+24$
$=-27+9 a+3 b+24$
$=9 a+3 b-3$
$\because x +3$ is a factor,
$\therefore$ Remainder $=0$,
$\therefore 9 a+3 b-3=0$
$\Rightarrow 3 a + b -1=0 \quad \ldots$ (Dividing by $3$ )
$\Rightarrow 3 a + b =1$
Again Let $x-4=0$,
then $x=4$
Substituting the value of $x$ in $f(x)$
$f(x)=(4)^2+a(4)^2-b(4)+24$
$=64+16 a-4 b+24$
$=16 a-4 b+88$
$\because x-4$ is a factor
$\therefore$ Remainder $=0$
$16 a-4 b+88=0$
$\Rightarrow 16 a-4 b=-88$
...(Dividing by 4)
$\Rightarrow 4 a - b =-22$
Adding (i) and (ii)
$7 a=-21,$
$\Rightarrow a=-3$
Substituting the value of $a$ in (i)
$3(-3)+b=1$
$\Rightarrow-9+b=1$
$\Rightarrow b=1+9=10$
$\therefore a=-3, b=10$
Now $f(x)$ will be
$f(x)=x^3-3 x^2-10 x+24$
$\because x+3$ and $x-4$ are factors of $f(x)$
$\therefore$ Dividing $f(x)$ by $(x+3)(x-4)$

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Question 75 Marks

If $x^3-2 x^2+p x+q$ has a factor $(x+2)$ and leaves a remainder 9 , when divided by $(x+1)$, find the values of $p$ and $q$. With these values of $p$ and $q$, factorize the given polynomial completely.

Answer

$f(x)=x^3-2 x^2+p x+q$
$(x+2)$ is a factor
$ f(-2)=(-2)^3-2(-2)^2+p(-2)+q$
$=-8-2 \times 4-2 p+q$
$=-8-8-2 p+q$
$=-16-2 p+q$
$\because(x+2)$ is a factor of $f(x)$
$\therefore f (-2)=0$
$\Rightarrow-16-2 p+q=0$
$\Rightarrow 2 p-q=-16$
Again, let $x+1=0$,
then $x=-1$
$f(-1)=(-1)^3-2(-1)^2+p(-1)+q$
$=-1-2 \times 1-p+q$
$=-1-2-p+q$
$=-3-p+q$
$\because$ Remainder $=9$, then
$-3-p+q=9$
$\Rightarrow-p+q=9+3=12$
$-p+q=12$
Adding (i) and (ii)
$p=-4$
Substituting the value of $p$ in (ii)
$-(-4)+q=12$
$4+q=12$
$\Rightarrow q=12-4=8$
$\therefore p=-4, q=8$
$\therefore f(x)=x^3-2 x^2-4 x+8$
Dividing $f(x)$ by $(x+2)$, we get
$f(x)=(x+2)\left(x^2-4 x+4\right)$
$=(x+2)\left\{(x)^2-2 x \times(-2)+(2)^2\right\}$
$=(x+2)(x-2)^2$

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Question 85 Marks

If $(2 x+1)$ is a factor of both the expressions $2 x^2-5 x+p$ and $2 x^2+5 x+q$, find the value of $p$ and $q$. Hence find the other factors of both the polynomials.

Answer

Let $2 x+1=0$, then $2 x=-1$
$x=-\frac{1}{2}$
Substituting the value of $x$ in
$ f(x)=2 x^2-5 x+p$
$f\left(-\frac{1}{2}\right)=2\left(\frac{-1}{2}\right)-5\left(\frac{-1}{2}\right)+p$
$=2 \times \frac{1}{4}+\frac{5}{2}+p$
$=\frac{1}{2}+\frac{5}{2}+p$
$=3+p$
$\because 2 x+1$ is the factor of $p(x)$
$\therefore$ Remainder $=0$
$ \Rightarrow 3+ p =0$
$\Rightarrow p =-3$
Again substituting the value of $x$ in $q(x)$
$q( x )=2 \times 2+5 x + q$
$q\left(-\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)^2+5\left(-\frac{1}{2}\right)+q$
$=2 \times \frac{1}{4}-\frac{5}{2}+q$
$=\frac{1}{2}-\frac{5}{2}+q$
$=-\frac{4}{2}+q$
$= q -2$
$2 x +1$ is the factor of $q ( x )$
Remainder $=0$
$ \Rightarrow q -2=0$
$\Rightarrow q =2$
Hence $p=-3, q=2$
Now (i) $: 2 x+1$ is the factor of $p(x)$
$=2 x^2-5 x-3$
$\therefore$ Dividing $p ( x )$ by $2 x +1$,

$\therefore 2x^2 + 5x + 2 = (2x + 1)(x + 2).$

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Mathematics [5 marks sum]

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