Easy Algebra Step-by-Step

128 Easy Algebra Step-by-Step

Your task when factoring 6x^2 + 23 x+ 20 is to reverse the FOIL process

to obtain 6x^2 + 23 x + 20 = (2x + 5)(3x + 4) in factored form. As you work

through the problems, you will fi nd it useful to know the following: If the

fi rst and last terms of a factorable quadratic trinomial are positive, the signs

of the second terms in the two binomial factors of the trinomial have the

same sign as the middle term of the trinomial.

Problem Factor by trial and error.

a. x^2 + 9 x + 14

b. x^2 − 9 x + 14

c. x^2 + 5 x − 14

d. x^2 − 5 x − 14

e. 3 x^2 + 5 x − 2

f. 4 x^2 − 11 x − 3

Solution

a. x^2 + 9x + 14

Step 1. Because the expression has the form ax^2 + bx + c, look for two bino-

mial factors.

x^2 + 9 x+ 14 = ( )( )

Step 2. x^2 is the fi rst term, so the fi rst terms in the two binomial factors must

be x.

x^2 + 9 x+ 14 = (x )(x )

Step 3. 14 is the last term, and it is positive, so the last terms in the two

binomial factors have the same sign as 9, with a product of 14 and a

sum of 9. Try 7 and 2 and check with FOIL.

xx^2 + 919 xx 4 =?()xx+ )()(xx+

Check: ()x+ )(()xx+ =xx^22 ++++ 72 xx++++ 2 xxx+ 1414 =+x 91 xx + 4

Correct



Step 4. Write the factored form.

xx

(^2) + 991 xx 4 ()xxx 7 ()x+ 2