If the leading coefficient after factoring the GCF (if possible) is a ≠ 1, then use the “bottoms up” method.

“Bottoms Up” Factoring

Find the factors that multiply to equal ac and add to equal b.

Write each factor as a fraction with the leading coefficient, a, as the denominator.

Reduce the fraction.

If a number remains in the denominator after reducing, bring the bottom up to become the coefficient of the parenthesis.

Factoring with a ≠ 1

f(x) = 2x2 – 11x + 12

(x – )(x – )

ac = 24, factors that multiply to equal 24

b = -4

-1 and -24 -1 + (-24) = -25

-2 and -12 -2 + (-12) = 14

-3 and -8 -3 + (-8) = -11

-4 and -6 -4 + (-6) = -10

(x – 3)(x – 8)

The factors are -3 and 8

Write factors as fractions over the leading coefficient: 2

Reduce

(x – )(x – 4)

(2x – 3)(x – 4)

Bring the “bottom” up

f(x) = 3x2 + 8x + 4

(x + )(x + )

(x + 2)(x + 6)

The factors are 2 and 6

Write factors as fractions over the leading coefficient: 2

Reduce

(x + )(x + 2)

(3x + 2)(x + 2)

Bring the “bottom” up

ac = (3)(4) = 12

b = 8

f(x) = 36x2 – 33x + 6

3(x – )(x – )

3(x – 3)(x – 8)

The factors are -3 and 8

Write factors as fractions over the leading coefficient: 2

Reduce

3(x – )(x – 4)

3(2x – 3)(x – 4)

Bring the “bottom” up

Factor out the GCF

f(x) = 3(12x2 – 11x + 2)

Determine the equation of a quadratic function that has roots of -3 and .

x = -3 x =

x + 3 = 0 5x = 2

(x + 3)(5x – 2)

5x2 + 13x – 6

g(x) = 5x2 + 13x – 6

Write the zeros as solutions for two equations.

Rewrite each equation so that it equals 0.

Multiply the binomials.

Name the function.

These two equations will represent the parenthesis had you factored the function.

x + 3 = 0 5x – 2 = 0