Question

Can anyone please help me solving these questions...?

Answer 1

Hey Abha here is your answer..... ...✌✌
.

.
.
.
1))) I have the attached the file for this answer.
.
.
.
.
.
.
.2))))))


GIVEN: p(x) = ax^4+2x^3–3x²+bx

g(x) = x²-4

Where p(x) & g(x) are polynomials in variable ‘x'

& here g(x) is a factor of p(x)

That means p(x) is exactly divisible by g(x)

Or p(x) is exactly divisible by factors of g(x)

Now, factors of g(x) = x²-4 = (x+2)(x-2)

So, p(x) is exactly divisible by (x+2) & (x-2)

That means, if p(x) is divided by (x+2) & then by (x-2) , the remainder has to be zero.

So now we find out the remainder in each case by remainder theorem:

If p(x) ÷(x+2) , the ramainder = p(-2)

ie, p(x)= ax^4+2x^3–3x²+bx is divided by (x+2)

Remainder= p(-2)= 16a - 16 -12 -2b =0………(1)

& if p(x)= ax^4+2x^3–3x²+bx is divided by (x-2)

Remainder= p(2)= 16a+16–12+2b=0……….(2)

eq(1) +eq(2)

=> 32a-24=0

=>32a =24

So a= 24/32 = 3/4………(3)

Now, eq(1) _ eq(2)

=> -32 -4b =0

=> 4b = -32

=> b= -8…………(4)

ANS a= 3 /4

b= —8
.
.
...
.
.
.hope it helps.. ✌✌