Answer:
(x - 4) is a factor of x^3 - 13x - 12 and x^3 - x^2 - 10x - 8
Explanation:
To solve this we can use the remainder theorem, which states that when a polynomial is divided by (x - a), then substituting a (making the brackets equal to zero) into the equation will give us the remainder of this equation.
So for this problem first we need to find a:
x - 4 = 0
x = 4
∴ a = 4
Now just substitute 4 into all the x's in each equation. If we are left with zero at the end, we know we have a remainder of zero, and therefore (x - 4) is a factor of that equation. Hence:
First equation:
y = x^3 - 13x - 12
y = 4^3 - 13(4) - 12
y = 64 - 52 - 12
y = 0
∴ (x - 4) is a factor
Second Equation:
y = x^3 + 8x^2 + 19x + 12
y = 4^3 + 8(4^2) + 19(4) + 12
y = 64 + 128 + 76 + 12
y = 280
∴ (x - 4) is not a factor
Third Equation:
y = x^3 + 6x + 5x - 12
y = 4^3 + 6(4) + 5(4) - 12
y = 64 + 24 + 20 - 12
y = 96
∴ (x - 4) is not a factor
Fourth Equation:
y = x^3 - x^2 - 10x - 8
y = 64 - 16 - 40 - 8
y = 0
∴ (x - 4) is a factor
Fifth Equation:
y = x^2 - 4
y = 16 - 4
y = 12
∴ (x - 4) is not a factor
(x - 4) is a factor of x^3 - 13x - 12 and x^3 - x^2 - 10x - 8
Hope that helps!