When the polynomial 4x^4 - 37x^2 + 9 is factored completely into linear factors of the for… — ACT

When the polynomial $4x^4 - 37x^2 + 9$ is factored completely into linear factors of the form $ax + b$ , where $a$ and $b$ are integers and $a > 0$ , which of the following expressions represents the sum of these linear factors?

A
$4x$
B
$6x - 8$
$6x$ Answer
D
$6x^4$
E
$6x - 6$

Answer

The sum of the linear factors is

6x

Factoring the polynomial

4x^4 - 37x^2 + 9

by substituting

u = x^2

yields

(4u - 1)(u - 9)

, which becomes

(4x^2 - 1)(x^2 - 9)

. Applying the difference of squares identity to both terms results in the four linear factors

(2x - 1)

(2x + 1)

(x - 3)

, and

(x + 3)

. The sum of these factors is

6x

Step-by-Step Solution

Substitute

u = x^2

to rewrite the quartic polynomial as a quadratic expression.

4u^2 - 37u + 9

This simplifies the polynomial from degree 4 to degree 2, making it easier to factor.

Factor the quadratic expression by grouping or finding two numbers that multiply to

36

and add to

-37

(4u - 1)(u - 9)

The numbers are

-36

and

-1

. Rewriting and grouping gives

4u(u - 9) - 1(u - 9) = (4u - 1)(u - 9)

Substitute

x^2

back in place of

u

and factor the resulting difference of squares binomials.

(2x - 1)(2x + 1)(x - 3)(x + 3)

Since

4x^2 - 1 = (2x)^2 - 1^2

and

x^2 - 9 = x^2 - 3^2

, both binomials can be factored completely using the difference of squares identity.

Sum the four linear factors.

6x

Combining the like terms gives

(2x - 1) + (2x + 1) + (x - 3) + (x + 3) = (2x + 2x + x + x) + (-1 + 1 - 3 + 3) = 6x + 0 = 6x

Key Concept

Factoring quartic polynomials using quadratic substitution and the difference of squares identity.

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