1.3. QUADRATIC EQUATIONS 21
So, for example, x
2
3x C 2 D .x 1/.x 2/ uses the facts that .1/.2/ D 2 and 1 2 D
3 to tell us that the factors are .x 1/ and .x 2/, which makes the roots 1 and 2. If .x a/
is a factor of a quadratic, then a is a root of the quadratic.
Example 1.45 Factor x
2
7x C 12 D 0, but first plug in x D 3.
Solution:
We plug in x D 3 and get that 9 21 C 12 D 0. So plugging in 3 makes the quadratic
zero. is means:
x
2
7x C 12 D .x 3/.x‹/
But 12/3=4 so ‹ D 4 is a good guess. Let’s check:
.x 3/.x 4/ D x
2
3x 4x C 12 D x
2
7x C 12;
and we have the correct factorization.
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e preceding example used a trick—plugging in 3—to locate one of the factors of the quadratic.
Let’s make this a formal rule in Knowledge Box 1.10. In general we don’t plug in 3, we plug in
things that might work. Factors of the constant terms are usually good guesses—remember that
they can be positive or negative.
Knowledge Box 1.10
Suppose that f .x/ is a quadratic equation and that f .c/ D 0 for some
number c. en .x c/ is a factor of f .x/.
Example 1.46 Factor x
2
2x 8 D 0.
Solution:
e number 8 factors into 1 8 and 2 4. Since 4 2 D 2 and the middle term is 2x,
the factorization is probably .x ˙ 2/.x ˙ 4/. Since we have 8, one needs to be positive, and
the other needs to be negative. To get a 2x we need to subtract 4x and add 2x. is means
the factorization must be:
x
2
2x 8 D .x 4/.x C 2/
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