5 Mark Question

Question 15 Marks

Find the values of p and q so that$ x^4 + px^3 + 2x^3 - 3x + q$ is divisible by $(x^2 - 1).$

Answer

Let $g(x)=x^2-1$, $p ( x )= x ^4+ px { }^3+2 x ^3-3 x + q$
Let $g ( x )=0$
$\Rightarrow x ^2-1=0$
$\Rightarrow x ^2=1$
$x= \pm \sqrt{1}= \pm 1$
$\because g(x)$ is a factor of $p(x)$
$\therefore p(1)=0, p(-1)=0$
$p(1)=1^4+p(1)^3+2(1)^2-3(1)+q=0$
$\Rightarrow 1+p+2-3+q=0$
$\Rightarrow p+q=0$
$\Rightarrow p=-q \ldots(1) p(-1)=(-1)^4+p(-1)^3+2(-1)^2-3(-1)+q=0$
$\Rightarrow 1+p(-1)+2+3+q=0$
$\Rightarrow 6-p+q=0$
$\Rightarrow p=6+q \ldots(2) \text { Equating (1) and }(2)-q=6+q$
$\Rightarrow-q-q=6$
$\Rightarrow-2 q=6$
$\Rightarrow q=\frac{-6}{2}=-3$
$\therefore p=6+(-3)=6-3=3$
$\therefore p=3, q=-3$

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Question 25 Marks

Find the values of $a$ and $b$ so that $(x+1)$ and $(x-1)$ are factors of $x^4+a x^3-3 x^2+2 x+b$.

Answer

Let, $x+1=0 x=-1$
$\because(x+1)$ is a factor of
$f(x)=x^4+a x^3-3 x^3-3 x^2+2 x+b$
$\therefore f(-1)=0$
$f(-1)=(-1)^4+a(-1)^3-3(-1)^2+2(-1)+b=0$
$\Rightarrow 1- a -3-2+ b =0$
$\Rightarrow-4- a + b =0 \Rightarrow b=4+ a$
Let, $x-1=0 x =1$
$\because(x-1)$ is a factor of $f(x)$
$\therefore f(1)=0$
$f(1)=1^4+a(1)^3-3(1)^2+2(1)+b=0$
$\Rightarrow 1+ a -3+2+ b =0$
$\Rightarrow a + b =0$
$\Rightarrow b =- a .$. (2) Equating (1) and (2)
$4+a=-a$
$\Rightarrow 4=-2 a$
$\Rightarrow a =-2$
$\therefore b=-a=-(-2)=2$ from (2)
$\therefore a=-2, b=2$

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Question 35 Marks

Show that $(x+4),(x-3)$ and $(x-7)$ are factors of $x^3-6 x^2-19 x+84$.

Answer

$x+4=0$
$\Rightarrow x=-4 f(x)$
$=x^3-6 x^2-19 x+84 f(-4)$
$=(-4)^3-6(-4)^2-19(-4)+84$
$=-64-6(16)-19(-4)+84$
$=-64-96+76+84=-160+160$
$=0$
Let, $x-3=0$
$\Rightarrow x=3 f(3)$
$=(3)^3-6(3)^2-19(3)+84$
$=27-6(9)-57+84$
$=27-54-57+84$
$=111-111=0$
Let, $x-7=0$
$\Rightarrow x=7 f(7)$
$=7^3-6(7)^2-19(7)+84$
$=343-6(49)-133+84$
$=343-294-133+84=427-427$
$=0$
$\because f(-4)=0, f(3)=0, f(7)=0$
$\therefore$ By factore theoram $( x +4),( x -3)$ and $( x -7)$ are factors of $f ( x )$.

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Question 45 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
$f(x) = x^3 - 6x^2 + 2x - 4, g(x) = 1 - 2x$

Answer

Here, $f(x) = x^3 - 6x^2 + 2x - 4 g(x) = 1 - 2x$
From, the remainder theorem when $f(x)$ is divided by
$\text{g(x)}=-2\Big(\text{x}-\frac{1}{2}\Big),$
the remainder will be equal to $\text{f}\Big(\frac{1}{2}\Big)$
Let, $g(x) = 0$
$\Rightarrow 1 - 2x = 0$
$\Rightarrow -2x = -1$
$\Rightarrow\ \text{x}=\frac{1}{2}$
Substitute the value of x in f(x)$\text{f}\Big(\frac{1}{2}\Big)=\Big(\frac{1}{2}\Big)^3-6\Big(\frac{1}{2}\Big)^2+2\Big(\frac{1}{2}\Big)-4$
$=\Big(\frac{1}{8}\Big)-8\Big(\frac{1}{4}\Big)+2\Big(\frac{1}{2}\Big)-4$
$=\Big(\frac{1}{8}\Big)-\Big(\frac{1}{2}\Big)+1-4$
$=\frac{1}{8}-\Big(\frac{1}{2}\Big)-3$
Taking L.C.M.$=\frac{1-4+8-32}{8}$
$=\frac{1-36}{8}$
$=\frac{-35}{8}$
Therefore, the remainder is $\Big(\frac{-35}{8}\Big).$

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Question 55 Marks

If polynomials $ax^3 + 3x^2 - 3$ and $2x^3 - 5x + a$ when divided by $(x - 4)$ leave the remainders as $R_1$ and $R_2$ respectively.
Find the values of a in each of the following cases, if

$R_1 = R_2$
$R_1 + R_2 = 0$
$2R_1 − R_2 = 0$

Answer

Here,
The polynomials are
$f(x)=a x^3+3 x^2-3$
$p(x)=2 x^3-5 x+a$
Let,
$R_1$ is the remainder when $f(x)$ is divided by $x-4$
$\Rightarrow R_1=f(4)$
$\Rightarrow R_1=a(4)^3+3(4)^2-3$
$=64 a+48-3$
$=64 a+45 \ldots(1)$
Now,
Let,
$R_2$ is the remainder when $p(x)$ is divided by $x-4$
$\Rightarrow R_2=p(4)$
$\Rightarrow R_2=2(4)^3-5(4)+a$
$=128-20+a$
$=108+a \ldots(2)$
1. Given,
$R_1=R_2$
$\Rightarrow 64 a+45=108+a$
$\Rightarrow 63 a=63$
$\Rightarrow a=1$
2. Given,
$R_1+R_2=0$
$\Rightarrow 64 a+45+108+a=0$
$\Rightarrow 65 a+153=0$
$\Rightarrow a=\frac{-153}{65}$
3. Given,
$2 R_1-R_2=0$
$\Rightarrow 2(64 a+45)-108-a=0$
$\Rightarrow 128 a+90-108-a=0$
$\Rightarrow 127 a-18=0$
$\Rightarrow a=\frac{18}{127}$

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Question 65 Marks

Find $\alpha$ and $\beta,$, if x + 1 and x + 2 are factors of $\text{x}^3+3\text{x}^2-2\alpha\text{x}+\beta.$

Answer

Let x + 1 = 0 ⇒ x = -1$\because$ x + 1 is a factor of $\text{f(x)}=\text{x}^3+3\text{x}^2-2\alpha\text{x}+\beta$
$\therefore\ \text{f}(-1)=0$
$\text{f}(-1)\Rightarrow\ (-1)^3+3(-1)^2-2\alpha(-1)+\beta=0$
$\Rightarrow\ -1+3+2\alpha+\beta=0$
$\Rightarrow\ 2+2\alpha+\beta=0$
$\Rightarrow\ \beta=-2\alpha-2\dots(1)$
Let, x + 2 = 0 ⇒ x = -2$\because$ x + 2 is a factor of f(x)
$\therefore$ f(-2) = 0
$\text{f}(-2)=(-2)^3+3(-2)^2-2\alpha(-2)+\beta=0$
$\Rightarrow\ -8+12+4\alpha+\beta=0$
$\Rightarrow\ 4+4\alpha+\beta=0$
$\Rightarrow\ \beta=-4-4\alpha\dots(2)$
Equating equations (1) and (2)$-2\alpha-2=-4-4\alpha$
$\Rightarrow\ -2\alpha+4\alpha=-4+2$
$\Rightarrow\ 2\alpha=-2$
$\Rightarrow\ \alpha=-1$
Substituting $\alpha=-1$ in equation (1)$\beta=-2(-1)-2=2-2=0$
$\therefore\ \alpha=-1,\beta=0$

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