Consider this unconventional method of factoring trinomials
ax2 + bx + c where
Jim’s Weird Method for Factoring
Trinomials - leading coefficient not 1.
Factor:
1. Multiply the leading coefficient (the
number in front of x2 ) times the constant
term (the last term). Remove the leading
coefficient.
(Looks like an illegal algebraic move! - but
is just a easier-to-see shorthand for what is
really happening.)
2. Factor this new trinomial.
3. Now undo step one by dividing the
constant terms by the same number used to
multiply the constant term in step #1.
4. Simplify.
5. These are the factors of the original
trinomial.
6. Check by multiplying the two binomials.
8
3
(𝑥 + ) (𝑥 + )
2
2
(x+4)(2x+3)
2𝑥 2 + 11𝑥 + 12
Does it always work? Yes!
Why does this work?
See next page!!
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The key to understanding this method is to realize that these
steps are actually a shorthand for a more complex process of
multiplication and replacement.
Take one more look at the whole process. The method given is
simply a shorthand version of the following procedure.
1. Original Trinomial
2𝑥 2 + 11𝑥 + 12
2. Multiply the entire expression
time the leading coefficient:
2(2𝑥 2 + 11𝑥 + 12)
3. Distribute:
4𝑥 2 + 22𝑥 + 24
4. Re-write:
(2𝑥)2 + (2𝑥 )11 + 24
5. Replace 2x with another
𝑚2 + 11𝑚 + 24
variable, such m, where 2x = m:
(𝑚 + 8)(𝑚 + 3)
6. Factor:
(2𝑥 + 8)(2𝑥 + 3)
7. Replace m with 2x:
8. Factor a 2 from the first
2(𝑥 + 4)(2𝑥 + 3)
binomial:
9. These factors now equal what we started with in step # 2, which is
TWICE the original trinomial in step #1:
2(2𝑥 2 + 11𝑥 + 12) = 2(𝑥 + 4)(2𝑥 + 3)
10. So divide both sides by 2 : 2𝑥 2 + 11𝑥 + 12 = (𝑥 + 4)(2𝑥 + 3)
11. And you are back to the original trinomial and its factored form:
2𝑥 2 + 11𝑥 + 12
(𝑥 + 4)(2𝑥 + 3)
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