Question

A polynomial has been factored below, but some constants are missing.

2x³8x²-24x = ax(x + b)(x + c)
What are the missing values of a, b, and c?
O a 2, b--2 and c = 6
O a=2, b=2 and c=-6
O a 1, b -4 and c=6
O a 1, b
4 and c =-6
Submit
Pass

Answer 1

`a=2, b=2 \;and\; c= - 6`

Explanation:

The given polynomial is
ax³ - 8x² -24x

Expand ax(x+b)(x+c) and compare

(x + b)(x + c) using the FOIL method is

x²+cx+bx+bc = x² + (b + c)x + bc

ax(x+b)(x+c) = ax(x² + (b + c)x + bc)

= ax³ + a(b+c)x² + abcx

The coefficient of a must be 2

abc must be -24 and since a = 2, 2bc = -24
bc = -24/2 = -12

This means b = -12/c

a(b+c) = -8

Since a = 2, 2(b+c) = -8

b + c = -8/2 = -4

Substitute b = -12/c in b+c = -4 to get
-12/c + c = -4

Multiply by c throughout
-12 + c² = -4c

c² + 4c -12 = 0

Factor
(c + 6)(c - 2) = 0

So c = -6 or c = 2

The only choice which has c = - 6 is the second choice
So c = -6 and b = -12/-6 = 2

Answer
a=2, b=2 and c= - 6

We could verify by plugging back the values into ax(x + b) ( x + c)

= ax³+ax² (c+b) +abcx to get
2x³ + 2x²(-6 + 2) + 2(2)(-6)x
= 2x³ +2x² (-4) + -24x

= 2x³ - 8x - 24x which is the given polynomial