Factoring with a ≠ 1
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.
f(x) = 2x2 – 11x + 12
The factors are -3 and 8
(x – 3)(x – 8)
Write factors as fractions
over the leading coefficient: 2
(x –
3
)(x
2
–
8
)
2
Reduce
3
2
(x – )(x – 4)
Bring the “bottom” up
(2x – 3)(x – 4)
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
f(x) = 3x2 + 8x + 4
The factors are 2 and 6
(x + 2)(x + 6)
Write factors as fractions
over the leading coefficient: 2
(x +
2
)(x
3
+
6
)
3
Reduce
2
3
(x + )(x + 2)
Bring the “bottom” up
(3x + 2)(x + 2)
ac = (3)(4) = 12
b=8
f(x) = 36x2 – 33x + 6
Factor out the GCF
f(x) = 3(12x2 – 11x + 2)
The factors are -3 and 8
3(x – 3)(x – 8)
Write factors as fractions
over the leading coefficient: 2
3(x –
3
)(x
2
–
8
)
2
Reduce
3
2
3(x – )(x – 4)
Bring the “bottom” up
3(2x – 3)(x – 4)
Determine the equation of a quadratic function
𝟐
that has roots of -3 and .
𝟓
x = -3
x=
2
5
x+3=0
5x = 2
x+3=0
5x – 2 = 0
(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.
These two equations will represent the
parenthesis had you factored the function.
Multiply the binomials.
Name the function.