Help 4 factor theorems:?

Question

Can anyone solve:::::::
The polynomial 2x^3 + x^2 + px - 4 has a factor x - 2.Find p. Show that 2x + 1 is also a factor. And deduce the third factor.

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

The polynomial 2x^3 + x^2 + px - 4 has a factor x - 2.Find p. Show that 2x + 1 is also a factor. And deduce the third factor.

if x-2 is a factor, then 2(2)^3 +2^2 + p(2) -4 =0, i.e.
16+4+2p - 4 =0
2p = -16
p = -8

so the polynomial is
2x^3 + x^2 -8x - 4.
and
2x^3 + x^2 -8x - 4 = (2x + 1) (x - 2) (x + 2).

Answer 2

(1) Let
p(x) = 2x^3 + x^2 + px - 4

Since it is given that (x - 2) is a factor, it follows that 2 is a root, and
p(2) = 0. However, if we solve for p(2) directly using the function, it is also equal to
p(2) = 2(2)^3 + 2^2 + p(2) - 4
p(2) = 2(8) + 4 + 2p - 4
p(2) = 16 + 2p

Equating this to 0, we get
16 + 2p = 0
2p = -16, p = -8

Now that we have p = -8, we can write the actual function with p solved:

p(x) = 2x^3 + x^2 - 8x - 4

We can actually group this now:

p(x) = x^2 (2x + 1) - 4(2x + 1)
p(x) = (2x + 1) (x^2 - 4)
p(x) = (2x + 1) (x - 2) (x + 2)

Which shows that (2x + 1) is a factor, and that the third factor is (x + 2).

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

Since we know it is exactly divisible by x^2 - 3x - 4, that means it is exactly divisible by its factors; x^2 - 3x - 4 = (x - 4) (x + 1), so if (x - 4) and (x + 1) are both factors, it follows that

p(4) = 0, and p(-1) = 0. Let's solve for p(4) and p(-1). Note that we can immediately equate them to 0 thereafter.

p(4) = a(4)^3 + b(4)^2 - 5(4) + 2a = 0

64a + 16b - 20 + 2a = 0
66a + 16b = 20, which can be reduced to

33a + 8b = 10

p(-1) = a(-1)^3 + b(-1)^2 - 5(-1) + 2a = 0
-a + b + 5 + 2a = 0
a + b = -5

Two equations, two unknowns are:

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

I won't show you the steps of solving this system (that can be homework for you), but your final answer should be:

a = 2
b = -7

Therefore, p(x), formerly known as ax^3 + bx^2 - 5x + 2a is now
p(x) = 2x^3 - 7x^2 - 5x + 4

To get the third and final factor, we use synthetic long division with x^2 - 3x - 4. Again, I won't show the steps, but after the synthetic long division, you should obtain (2x - 1).

Therefore, the complete factorization of p(x) should be

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