Question

X + a is a factor of x³ + 8x² + 4ax - 3a

a) Show that a³ - 4a² + 3a = 0.
b) Find the possible values of a.

I have done question A, but I'm not sure on how to do question B.​

Answer 1

To find the values of a, factoring is used, yielding a simple quadratic equation. The factorization results in the values of a being 3 and 1, satisfying the condition that the equation has (x + a) as a factor.

Step-by-step explanation:

To solve for the possible values of a in question B, given that you have already shown a³ - 4a² + 3a = 0, you will need to factor the expression or use the quadratic formula if it is a quadratic equation in disguise. In this case, you can factor out an a, leading to:

a(a² - 4a + 3) = 0

Now, you have a simple quadratic equation a² - 4a + 3 = 0, which you can factor further:

(a - 3)(a - 1) = 0

Therefore, the possible values of a are 3 and 1. These are the roots of the quadratic equation, which satisfy the condition that when a is substituted back into the original cubic equation, x + a is a factor.

Answer 2

Step-by-step explanation:

If (x + a) is a factor of x³ + 8x² + 4ax - 3a, then we know that dividing x³ + 8x² + 4ax - 3a by x + a will give a quotient of x² + (8 - a)x + (4a - 3), and a remainder of 0.

We can write this as:

x³ + 8x² + 4ax - 3a = (x + a)(x² + (8 - a)x + (4a - 3))

Expanding the right-hand side, we get:

x³ + 8x² + 4ax - 3a = x³ + (8-a)x² + (4a-3)x + ax² + (8-a)ax + (4a-3)a

Collecting like terms, we get:

x³ + 8x² + 4ax - 3a = x³ + (a+8)x² + (5a-3) x + a(4a-3)

Comparing the coefficients of x², x and the constant term, we get the following equations:

a + 8 = 8 --> a = 0

5a - 3 = 0 --> a = 3/5

4a^2 - 3a = 0 --> a(4a-3) = 0

Therefore, we have:

a³ - 4a² + 3a = a(a² - 4a + 3) = a(a-1)(a-3) = 0

So, a can be 0, 1, or 3

Therefore, the possible values of a are 0, 1, and 3.