# Factor by using a quadratic pattern

Learn how to factor a higher-degree polynomial by using a quadratic pattern with these carefully chosen examples.

Example #1

Factor x4 + 7x2 + 6 by using a quadratic pattern

Step 1

Write x4 + 7x2 + 6 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.

Let y = x2 and substitute y for x2

x4 + 7x2 + 6 = (x2)2 + 7(x2) + 6

x4 + 7x2 + 6 = (y)2 + 7(y) + 6

Step 2

Factor y2 + 7y + 6

y2 + 7y + 6 = (y + ___ )(y + ___ )

To fill in the blank above, look for factors of 6 that will add up to 7.

6 × 1 = 6 and 6 + 1 = 7.

Fill in the blank in the expression above with 1 and 6.

y2 + 7y + 6 = (y + 1 )(y + 6)

Step 3

Substitute back to the original variable

(y + 1 )(y + 6) = (x2 + 1)(x2 + 6)

Example #2

Factor x4 - 4x2 - 45 by using a quadratic pattern

Step 1

Write x4 - 4x2 - 45 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.

Let y = x2 and substitute y for x2

x4 - 4x2 - 45 = (x2)2 - 4(x2) - 45

x4 - 4x2 - 45 = (y)2 - 4(y) - 45

Step 2

Factor y2 - 4y - 45

y2 - 4y - 45 = (y + ___ )(y + ___ )

To fill in the blank above, look for factors of -45 that will add up to -4.

-9 × 5 = 45 and -9 + 5 = -4.

Fill in the blank in the expression above with -9 and 5.

y2 - 4y - 45 = (y - 9)(y + 5)

Step 3

Substitute back to the original variable

(y - 9)(y + 5) = (x2 - 9)(x2 + 5)

Factor completely

(y - 9)(y + 5) = (x2 - 9)(x2 + 5) = (x - 3)(x + 3)(x2 + 5)

100 Tough Algebra Word Problems.

If you can solve these problems with no help, you must be a genius!

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