plz solve that?

Question

The expression ax^3 + bx^2 - 5x + 2a is exactly divisible by x^2 - 3x - 4 . Calculate the value of a and b.and factorise the expression completely.

Answer 1

Let p(x) = ax^3 + bx^2 - 5x + 2a, then,

Since it is given that p(x) is exactly divisible by x^2 - 3x - 4, then
it follows p(x) is divsible by its factors.
x^2 - 3x - 4 = (x - 4) (x + 1)

That means p(4) = 0 and p(-1) = 0.

But p(4) = a(4)^3 + b(4)^2 - 5(4) + 2a = 0
64a + 16b - 20 + 2a = 0
66a + 16b - 20 = 0
33a + 8b - 10 = 0
33a + 8b = 10

Similarly, if p(1) = 0, then
a + b - 5 + 2a = 0
3a + b = 5

Our two equations, two unknowns are:

33a + 8b = 10
3a + b = 5

Solving the system of equations would yield
a = -10/3, b = 15

So now that we have both values,
p(x) = (-10/3)x^3 + 15x^2 - 5x + 2(-10/3)
p(x) = (-10/3)x^3 + 15x^2 - 5x - 20/3

We now perform synthetic long division. I'll leave the rest to you.

Answer 2

when you perform the quotient, you obtain the following two eqs

13a + 3b = 5
7a + 2b = 0

in order to make the remain equal to 0. The result of the quotient is, by the way, ax + 3a + b, when you find the values of a & b, you use this result to the factorization. The solution of the system of equations is a = 2 and b = -7, so your polynomial is now

2 x^3 - 7 x^2 - 5 x +4

one of the factors is the one resulting from quotient and with the value of a & b from the system solution, i. e., 2 x -1. The other two factors are from the 2nd grade polynomial and they are (x - 4)(x + 1) to give finally

(2 x - 1)(x - 4)(x + 1)

By the way, the first answer that puggy gave is not exactly divisible by x^2 - 3x - 4