How do you factor quadratic trinomials?

Answer 1

ax2 + bx + c is the general formula for a quadratic trinomial.

If a = 1 it is easier. Find 2 integers whose product = c AND whose sum = b.

The 2 number must do both. call thos numbers m and n

The factors become (a+m)(a+n)

ex: x2 - x - 12

c = -12, has many factors -1 x 12, -2 x 6, -3 x 4, -4x3, -6x2, and -1x12

compare these to b = -1. the only ones that have a sum of -1 are -4 and 3

so the factors are (x-4)(x+3) --- answer

But if a is not equal to 1

step1. Multiply a and c find a new number, lets call it d. Use d instead of c like above to find factors and look for a sum equal to b. just like above but you are factoring d NOT c. lets call these factors m and n just like before.

step 2. Replace bx with mx + nx Now you have 4 terms instead of 3

ax2 + bx + c becomes ax2 + mx + nx + c group in twos using parenthesis

( ax2 + mx) + (nx + c)

step 3. Look at each group separately and get a GCF.

step 4, Look at the parenthesis after each above GCF. If they are not the same the original trinomial was prime (or you made a mistake). If they are the same then put the GCF's into a parenthesis multiplied by the parenthesis that was the same. This is the answer. This sounds complicated, but it is easier to do than to explain using a computer.

Ex: Factor 2x2 +11x +12

Step 1 --- 2x12 = 24, Factors are 1x24, 2x12, 3x8, 4x6 and some negatives which we don't need THIS TIME. Look for the sum that = b = 11 ... 3+8 = 11

Step 2 --- replace b ---- 2x2 +11x +12 = 2x2 + 8x + 3x +12

group ---- ( 2x2 + 8x) + ( 3x +12)

Step 3 --- GCF --- 2x(x + 4) + 3(x + 4) ----- notice parenthesis is same.

Step 4 ---- Both GCF's make a new group and same parenthesis is used once

(2x + 3)(x+4)

This is the answer.