5 Mark Question

Question 15 Marks

If $2$ and $0$ are the zeros of the polynomial $f(x) = 2x^3 - 5x^2 + ax + b$ then find the values of a and b.
Hint: $f(x) = 0$ and $f(0) = 0.$

Answer

It is given that 2 and 0 are the zeros of the polynomial
$f(x)=2 x^3-5 x^2+a x+b$
$\therefore f(2)=0$
$\Rightarrow 2 \times 2^3-5 \times 2^2+a \times 2+b=0$
$\Rightarrow 16-20+2 a+b=0$
$\Rightarrow-4+2 a+b=0$
$\Rightarrow 2 a+b=4 \ldots(1)$
Also, $f(0)=0$
$\Rightarrow 2 \times 0^3-5 \times 0^2+a \times 0+b=0$
$\Rightarrow 0-0+0+b=0$
$\Rightarrow b=0$
Putting $b=0$ in (1),
we get $2 a+0=4$
$\Rightarrow 2 a=4$
$\Rightarrow a=2$
Thus, the values of $a$ and $b$ are 2 and 0 , respectively.

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Question 25 Marks

If $p(x)=x^3+x^2-9 x-9$, find $p(0), p(3), p(-3)$ and $p(-1)$. What do you conclude about the zeros of $p(x)$ ? Is 0 a zero of $p ( x )$ ?

Answer

$p(x)=x^3+x^2-9 x-9 \ldots$
Putting $x=0$ in (1),
$\text { we get } p(0)=0^3+0^2-9 \times 0-9=-9 \neq 0$
Thus, $x=0$ is not a zero of $p(x)$.
Putting $x=3$ in (1),
we get $p(3)=3^3+3^2-9 \times 3-9$
$=27+9-27-9$
$=0$
Thus, $x=3$ is a zero of $p(x)$.
Putting $x=-3$ in (1),
we get $p(-3)=(-3)^3+(-3)^2-9 \times(-3)-9$
$=-27+9+27-9$
$=0$
Thus, $x=-3$ is a zero of $p(x)$. Putting $x=-1$ in (1),
we get $p(-1)=(-1)^3+(-1)^2-9 \times(-1)-9$
$=-1+1+9-9$
$=0$
Thus, $x=-1$ is a zero of $p(x)$.

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Question 35 Marks

If $\left(x^3+a x^2+b x+6\right)$ has $(x-2)$ as a factor and leaves a remainder 3 when divided by $(x-3)$, find the values of a and b.

Answer

Let $f(x)=\left(x^3+a x^2+b x+6\right)$
Now, by remainder theorem, $f(x)$ when divided by $(x-3)$ will leave a remainder as $f(3)$.
$\Rightarrow \text { So, } f(3)=3^3+a 3^2+b 3+6=3$
$\Rightarrow 27+9 a+3 b+6=3$
$\Rightarrow 9 a+3 b+33=3$
$\Rightarrow 9 a+3 b=3-33$
$\Rightarrow 9 a+3 b=-30$
$\Rightarrow 3 a+b=-10 \ldots$... $i$ ) Given that $(x-2)$ is a factor of $f(x)$.
By the Factor Theorem, $(x-a)$ will be a factor of $f(x)$ if $f(a)=0$ and therefore $f(2)=0$.
$\Rightarrow f(2)=2^3+a 2^2+b 2+6=0$
$\Rightarrow 8+4 a+2 b+6=0$
$\Rightarrow 4 a+2 b=-14$
$\Rightarrow 2 a+b=-7 \ldots \text { (ii) Subtracting (ii) from (i), }$
we get,
$\Rightarrow a=-3$
Substituting the value of $a=-3$ in (i),
we get, $3(-3)+b=-10$
$\Rightarrow-9+b=-10$
$\Rightarrow b=-10+9$
$\Rightarrow b=-1$
$\therefore a=-3 \text { and } b=-1$

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Question 45 Marks

Find the values of a and b so that the polynomial $(x^4 + ax^3 - 7x^2- 8x + b)$ is exactly divisible by $(x + 2)$ as well as $(x + 3).$

Answer

Let $f(x)=\left(x^4+a x^3-7 x^2-8 x+b\right)$
Now, $x+2=0$
$\Rightarrow x=-2 \text { and, }$
$\Rightarrow x+3=0$
$\Rightarrow x=-3$
By factor theorem, $(x+2)$ and $(x+3)$ will be factors of $f(x)$ if $f(-2)=0$ and $f(-3)=0$
$\therefore f(-2)=(-2)^4+a(-2)^3-7(-2)^2-8(-2)+b=0$
$\Rightarrow 16-8 a-28+16+b=0$
$\Rightarrow-8 a+b=-4$
$\Rightarrow 8 a-b=4 \ldots\left(\text { i) And, } f(-3)=(-3)^4+a(-3)^3-7(-3)^2-8(-3)+b=0\right.$
$\Rightarrow 81-27 a-63+24+b=0$
$\Rightarrow-27 a+b=-42$
$\Rightarrow 27 a-b=42 \ldots \text { (ii) Subtracting (i) from (ii), }$
we get, 19a = 38 So, $a=2$
Substituting the value of $a=2$ in (i),
we get $8 \times 2-b=4$
$\Rightarrow 16-b=4$
$\Rightarrow-b=-16+4$
$\Rightarrow-b=-12$
$\Rightarrow b=12$
$\therefore a=2 \text { and } b=12$

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Question 55 Marks

If $p(x) = 5 - 4x + 2x^2$, find:

$p(0)$
$p(3)$
$p(-2)$

Answer

i $p(x) = 5 - 4x + 2x^2$
$\Rightarrow p(0) = (5 - 4 \times 0 + 2 \times 0^2)$
$= (5 - 0 + 0)$
$= 5$
ii.$p(x) = 5 - 4x + 2x^2 p(3) = (5 - 4 \times 3 + 2 \times 3^2)$
$= (5 - 12 + 18)$
$= 11$
iii.$p(x) = 5 - 4x + 2x^2 \Rightarrow p(-2) = [(5 - 4 \times (-2) + 2 \times (-2)^2]$
$= (5 + 8 + 8)$
$= 21$

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Question 65 Marks

If $f(t) = 4t^2 - 3t + 6$, find:

f(0)
f(4)
f(-5)

Answer

i. $f(t)=4 t^2-3 t+6$
$\Rightarrow f(0)=\left(4 \times 0^2-3 \times 0+6\right)$
$= (0-0+6)$
$= 6$
ii. $f(t)=4 t^2-3 t+6$
$\Rightarrow f(4)=\left(4 \times 4^2-3 \times 4+6\right)$
$= (64-12+6)$
$= 58$
iii. $f(t)=4 t^2-3 t+6$
$\Rightarrow f(-5)=\left[\left(4 \times(-5)^2-3 \times(-5)+6\right)\right]$
$= (100+15+6)$
$= 121$

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Question 75 Marks

If both $(x - 2)$ and $\Big(\text{x}-\frac{1}{2}\Big)$ are factors of $px^2 + 5x + r$, prove that $p = r.$

Answer

Let f(x) = $px^2 + 5x + r$
Now, $(x - 2)$ is a factor of $f(x).$
$\Rightarrow f(2) = 0$
$\Rightarrow p(2)^2 + 5(2) + r = 0$
$\Rightarrow 4p + 10 + r = 0$
$\Rightarrow 4p + r = -10$ ...(i) Also, $\Big(\text{x}-\frac{1}{2}\Big)$ is a factor of f(x).
$\Rightarrow\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)+5\times\frac{1}{2}+\text{r}=0$
$\Rightarrow\frac{\text{p}}{4}+\frac{5}{2}+\text{r}=0$
$\Rightarrow\frac{\text{p}+10+4\text{r}}{4}=0$
$\Rightarrow\text{p}+4\text{r}+10=0$
$\Rightarrow\text{p}+4\text{r}=-10\ ...(\text{ii})$
From (i) and (ii), we have $4p + r = p + 4r$
$\Rightarrow 4p - p = 4r - r$
$\Rightarrow 3p = 3r$
$\Rightarrow p = r$

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Question 85 Marks

If $p(x)=x^3-3 x^2+2 x$, find $p(0), p(1), p(2)$. What do you conclude?

Answer

$p(x)=x^3-3 x^2+2 x \ldots(1)$
Putting $x=0$ in (1), we get
$p(0)=0^3-3 \times 0^2+2 \times 0=0$
Thus, $x=0$ is a zero of $p(x)$.
Putting $x=1$ in (1), we get
$p(1)=1^3-3 \times 1^2+2 \times 1$
$=1-3+2=0$
Thus, $x=1$ is a zero of $p(x)$.
Putting $x=2$ in (1), we get
$p(2)=2^3-3 \times 2^2+2 \times 2$
$=8-3 \times 4+4$
$=8-12+4=0$
Thus, $x=2$ is a zero of $p(x)$.

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Question 95 Marks

If p(y) = $4 - 3y -y^2 + 5y^3$, find:

$p(0)$
$p(2)$
$p(-1)$

Answer

$p(y) = 4 - 3y -y^2 + 5y^3$

$\Rightarrow p(0) = (4 + 3 \times 0 - 0^2 + 5 \times 0^3 )$

$= (4 + 0 - 0 + 0)$

$= 4$

$p(y) = 4 - 3y -y^2 + 5y^3$

$\Rightarrow p(2) = (4 + 3 \times 2 - 2^2 + 5 \times 2^3 )$

$= (4 + 6 - 4 + 40)$

$= 46$

$p(y) = 4 - 3y -y^2 + 5y^3$

$\Rightarrow p(-1) = [(4 + 3 \times (- 1)^2 + 5 \times (-1)^3 )]$

$= (4 - 3 - 1 - 5)$

$= -5$

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Question 105 Marks

Verify the division algorithm for the polynomials
$p(x) = 2x^4 - 6x^3 + 2x^2 - x + 2$ and $g(x) = x + 2.$

Answer

$p(x) = 2x^4 - 6x^3 + 2x^2 - x + 2$ and $g(x) = x + 2$

Quotient = $2x^3 - 10x^2 + 22x - 45$
Remainder = $92$
Verification:$ Divisor \times Quotient + Remainder$
$= (x + 2) \times (2x^3 - 10x^2 + 22x - 45) + 92$
$= x (2x^3 - 10x^2 + 22x - 45) + 2(2x^3 - 10x^2 + 22x - 45) + 92$
$= 2x^4 - 10x^3 + 22x^2 - 45x + 4x^3 - 20x^2 + 44x - 90 + 92$
$= 2x^4 - 6x^3 + 2x^2 - x + 2$
= Dividend Hence verified.

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Question 115 Marks

The polynomial $p(x)=x^4-2 x^3+3 x^2-a x+b$ when divided by $(x-1)$ and $(x+1)$ leaves the remainders 5 and 19 respectively.
Find the values of $a$ and $b$. Hence, find the remainder when $p(x)$ is divided by $(x-2)$.

Answer

Let: $p(x)=x^4-2 x^3+3 x^2-a x+b$
Now, When $p(x)$ is divided by $(x-1)$, the remainder is $p(1)$.
When $p(x)$ is divided by $(x+1)$, the remainder is $p(-1)$.
Thus, we have: $p(1)=\left(1^4-2 \times 1^3 \times+3 \times 1^2-a \times 1+b\right)$
$=(1-2+3-a+b)$
$=2-a+b \text { And, }$
$p(-1)=\left[(-1)^4-2 \times(-1)^3+3 \times(-1)^2-a \times(-1)+b\right]$
$=(1+2+3+a+b)=6+a+b$
Now, $2-a+b=5$
...(1) $6+a+b=19$
...(2) Adding (1) and (2),
we get: $8+2 b=24$
$\Rightarrow 2 b=18$
$\Rightarrow b=8$
By putting the value of $b$, we get the value of $a$, i.e., 5 .
$\therefore a=5 \text { and } b=8$
Now, $f(x)=x^4-2 x^3+3 x^2-5 x+8$
Also, When $p(x)$ is divided by $(x-2)$, the remainder is $p(2)$.
thus, we have: $p(2)=\left(2^4-2 \times 2^3+3 \times 2^2-5 \times 2+8\right)[a=5$ and $b=8]$
$=(16-16+12-10+8)=10$

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Question 125 Marks

What must be subtracted from $(x^4 + 2x^3 - 2x^2 + 4x + 6)$ so that the result is exactly divisible by $(x^2 + 2x - 3)$?

Answer

Let $p ( x )= x ^4+2 x ^3-2 x ^2+4 x +6$ and $q ( x )= x ^2+2 x -3$.
When $p(x)$ is divided by $q(x)$, the remainder is a linear expression in $x$.
So, let $r(x)=a x+b$ be subtracted from $p(x)$ so that $p(x)-r(x)$ is divided by $q(x)$.
$\text { Let } f(x)=p(x)-r(x)=p(x)-(a x+b)$
$=\left(x^4+2 x^3-2 x^2+4 x+6\right)-(a x+b)$
$=x^4+2 x^3-2 x^2+(4-a) x+6-b$
We have,
$q(x)=x^2+2 x-3$
$=x^2+3 x-x-3$
$=x(x+3)-1(x+3)$
$=(x+3)(x-1)$
Clearly, $(x+3)$ and $(x-1)$ are factors of $q(x)$.
Therefore, $f(x)$ will be divisible by $q(x)$ if $(x+3)$ and $(x-1)$ are factors of $f(x)$.
$\text { i.e., } f(-3)=0 \text { and } f(1)=0$
Consider, $f (-3)=0$
$\Rightarrow(-3)^4+2(-3)^3-2(-3)^2+(4-a)(-3)+6-b=0$
$\Rightarrow 81-54-18-12+3 a+6-b=0$
$\Rightarrow 3+3 a-b=0$
$\Rightarrow 3 a-b=-3 \ldots(i)$
And, $f(1)=0$
$\Rightarrow(1)^4+2(1)^3-2(1)^2+(4-a)(1)+6-b=0$
$\Rightarrow 1+2-2+4-a+6-b=0$
$\Rightarrow 11-a-b=0$
$\Rightarrow-a-b=-11 \ldots \text { (ii) }$
Subtracting (ii) from (i), we get
$4 a=8$
$\Rightarrow a=2$
Substituting $a=2$ in (i), we get
$3(2)-b=-3$
$\Rightarrow 6-b=-3$
$\Rightarrow b=9$
Putting the values of $a$ and $b$ in $r(x)=a x+b$, we get
$r(x)=2 x+9$
Hence, $p(x)$ is divisible by $q(x)$, if $r(x)=2 x+9$ is subtracted from it.

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