[4 marks sum]

Question 14 Marks

Given$ f(x) = ax^2 + bx + 2$ and $g(x) = bx^2 + ax + 1. If x – 2$ 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) + 4x^2 + 7x.$

Answer

$ f(x)=a x^2+b x+2$
$ g(x)=b x^2+a x+1$
$ x-2$ is $a$ 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 \ldots(\because x-2 s$ its factor $)$
$ \Rightarrow 2 a+b+1=0 \ldots (i) ($ Dividing by $2)$
Dividing $g(x) \text { by } x-2,$ remainder $=-15$
Let $x-2=0$
$\Rightarrow x=2$
$\therefore g(2)=b(2)^2+a x 2+1$
$ =4 b+2 a+1$
$\because$ Remainder is $-15$
$\therefore 4 b+2 a+1=-15$
$\Rightarrow 4 b+2 a+1+15=0$
$\Rightarrow 4 b+2 a+16=0$
$\Rightarrow 2 b+a+8=0 \ldots ($Dividing by $2) $
$ \Rightarrow a+2 b+8=0$
Multiplying $(i)$ by $2$ 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 24 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 $b=-2$
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 34 Marks

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

Answer

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

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Question 44 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 54 Marks

$f\ 2x^3 + ax^2 – 11x + b$ 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$
$\because$ 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 64 Marks

If $(x + 3)$ and $(x – 4)$ are factors of $x^3 + ax^2 – bx + 24,$ 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 $=0$,
$\therefore 9 a+3 b-3=0$
$\Rightarrow 3 a + b -1=0 \ldots($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 74 Marks

If $x^3 – 2x^2 + px + q$ 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 84 Marks

If $(2x + 1)$ is a factor of both the expressions $2x^2 – 5x + p$ and $2x^2 + 5x + q,$ 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 [4 marks sum]

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