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?

1 answer

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.