Question

A polynomial has been factored below, but some constants are missing.
2x^3-8x^2-24x=ax(x+b)(x+c)
What are the missing values of a, b, and c?

Answer 1

To find the missing values of a, b, and c, we can expand the right side of the equation and compare it to the given polynomial.

Expanding the right side:
ax(x+b)(x+c) = ax(x^2 + cx + bx + bc)
= ax^3 + acx^2 + abx^2 + abc

Setting this equal to the given polynomial:
2x^3 - 8x^2 - 24x = ax^3 + acx^2 + abx^2 + abc

Now we can compare coefficients to find the values of a, b, and c.

For the x^3 term:
2 = a, so a = 2

For the x^2 term:
-8 = ac + ab
-8 = 2c + 2b
c + b = -4

For the x term:
-24 = abc
Since a = 2, bc = -12

We need to find two numbers b and c whose product is -12 and whose sum is -4.

The factors of -12 that add up to -4 are -6 and 2, so b = -6 and c = 2.

Therefore, the missing values of a, b, and c are 2, -6, and 2.