Question

5. The polynomial p(x) = x3 + ax2 +bx -20 when divided by x -5 and x-3 leaves the remainders 0 and -2 respectively . Find the values of a and b.

6. If p and q are remainders when the polynomials x3 + 2x2 -5ax -7 and x3 + ax2 -12x + 6 are divided by x + 1 and x -2 respectively and if 2p + q = 6 , find a.

7. Without actual division , show that p(x) = (x -1)2a – x2a + 2x -1 is divisible by 2x3 -3x2 + x.​

Answer 1

ANSWER:5)

Given:

• p(x) = x^3 + ax^2 + bx - 20
• When p(x) divided by (x - 5), Remainder = 0
• When p(x) divided by (x - 3), Remainder = -2

To Find

• Value if a and b.

Solution:

We are given that,

$\implies p(x)=x^3+ax^2+bx-20$

We are also given that, when p(x) is divided by x - 5, the remainder is 0. So,

$\implies p(5)=0$

$\implies p(5)=(5)^3+a(5)^2+b(5)-20$

$\implies 0=125+25a+5b-20$

$\implies 0=105+25a+5b$

$\implies 25a+5b=-105$

$\implies 5a+b=-21$

$\implies b=-21-5a- - - -(1)$

We are also given that, when p(x) is divided by x - 3, the remainder is -2. So,

$\implies p(3)=-2$

$\implies p(3)=(3)^3+a(3)^2+b(3)-20$

$\implies -2=27+9a+3b-20$

$\implies -2=7+9a+3b$

$\implies 9a+3b=-9$

$\implies 3(3a+b)=-9$

$\implies 3a+b=-3$

From (1),

$\implies 3a+(-21-5a)=-3$

$\implies 3a-21-5a=-3$

$\implies -2a=-3+21$

$\implies -2a=18$

$\implies a=\dfrac{-18}{2}$

$\implies a=-9 - - - -(2)$

So, we had,

$\implies b=-21-5a$

From (2),

$\implies b=-21-5(-9)$

$\implies b=-21+45$

Hence,

$\implies b=24$

Therefore, the value of a and b is -9 and 24 respectively.

Answer:6)

Given:

• p(x) = x^3 + 2x^2 - 5ax - 7
• f(x) = x^3 + ax^2 - 12x + 6
• When, p(x) is divided by x + 1, Remainder is p.
• When, f(x) is divided by x - 2, Remainder is q.
• 2p + q = 6

To Find:

• Value of a

Solution:

We are given that,

$\implies p(x) = x^3+2x^2-5ax-7$

Also, when, p(x) is divided by x + 1, Remainder is p.

So,

$\implies p(-1)=p$

$\implies p(-1)=(-1)^3+2(-1)^2-5a(-1)-7$

$\implies p=-1+2(1)+5a-7$

$\implies p=-1+2-7+5a$

$\implies p=5a-6 - - - -(1)$

We are also given that,

$\implies f(x) = x^3 + ax^2 - 12x + 6$

Also, when, f(x) is divided by x - 2, Remainder is q.

So,

$\implies f(2)=q$

$\implies f(2)=(2)^3+a(2)^2-12(2)+6$

$\implies q=8+4a-24+6$

$\implies q=8-24+6+4a$

$\implies q=4a-10 - - - - (2)$

We also have,

$\implies 2p + q = 6$

From (1) & (2),

$\implies 2(5a-6) + (4a-10) = 6$

$\implies 10a-12+4a-10=6$

$\implies 10a+4a-12-10=6$

$\implies 14a-22=6$

$\implies 14a=6+22$

$\implies 14a=28$

$\implies a=\dfrac{28}{14}$

So,

$\implies\bf a=2$

Therefore, the value of a is 2.

Answer:6)

To Prove:

• p(x) = (x -1)$^{2a}$ - x$^{2a}$ + 2x - 1 is divisible by 2x^3 - 3x^2 + x.

Solution:

We are given that,

$\implies p(x)=(x-1)^{2a}-x^{2a}+2x-1$

Let us assume that,

$\implies 2x^3-3x^2+x=f(x)$

That is,

$\implies f(x)=2x^3-3x^2+x$

Now, we will first simplify f(x)

$\implies f(x)=2x^3-3x^2+x$

$\implies f(x)=x(2x^2-3x+1)$

$\implies f(x)=x(2x^2-2x-x+1)$

$\implies f(x)=x(2x(x-1)-1(x-1))$

$\implies f(x)=x(2x-1)(x-1)$

So, for f(x),

$\implies x(2x-1)(x-1)=0$

Hence,

$\implies x=0,1,\frac{1}{2}$

Now, for p(x) to be divisible by f(x), all the zeroes of f(x) should be zeroes of p(x).

Now, we know that zeroes of f(x) are: 0, 1 and 1/2.

Therefore,

$\implies p(0)=p(1)=p\left(\dfrac{1}{2}\right)=0$

Taking the zeroes one by one,

First, p(0),

$\implies p(0)=(0-1)^{2a}-0^{2a}+2(0)-1$

$\implies p(0)=(-1)^{2a}-0+0-1$

$\implies p(0)=(-1)^{2a}-0-0-1$

As, 2a is divisible by 2, it is even.

And, even power of -1 = 1. So,

$\implies p(0)=(-1)^{2a}-1$

$\implies p(0)=1-1$

$\implies\bf p(0)=0$

Taking p(1) now,

$\implies p(1)=(1-1)^{2a}-1^{2a}+2(1)-1$

$\implies p(1)=(0)^{2a}-1+2-1$

$\implies p(1)=0-1+2-1$

$\implies p(1)=2-2$

$\implies\bf p(1)=0$

Now, we'll take p(1/2).

$\implies p\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}-1\right)^{2a}-\left(\dfrac{1}{2}\right)^{2a}+2\left(\dfrac{1}{2}\right)-1$

$\implies p\left(\dfrac{1}{2}\right)=\left(\dfrac{-1}{2}\right)^{2a}-\left(\dfrac{1}{2}\right)^{2a}+1-1$

$\implies p\left(\dfrac{1}{2}\right)=(-1)^2\left(\dfrac{1}{2}\right)^{2a}-\left(\dfrac{1}{2}\right)^{2a}+1-1$

$\implies p\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{2a}-\left(\dfrac{1}{2}\right)^{2a}+0$

Cancelling (1/2)^(2a),

So,

$\implies\bf p\left(\dfrac{1}{2}\right)=0$

As,

$\implies\bf p(0)=p(1)=p\left(\dfrac{1}{2}\right)=0$

Therefore, p(x) is divisible by f(x).

HENCE PROVED!!