Factoring a Non-perfect Square Trinomial
Trinomials with Leading Coefficient One
Consider (x + a)(x + b) = x² + bx + ax + ab
= x² + ax + ba + ab Combine like terms.
= x² + (a + b)x + ab
Note the trinomial given on the right has last term ab and the coefficient
of the middle term is the sum of a and b. To factor a trinomial with
leading coefficient one we need only to find two numbers a and b whose
product is ab and whose sum is the numerical coefficient of the middle
term.
Example: Factor x² + 3x - 28.
Factors of -28: Sum of factors:
1(-28) -27
-1(28) 27
2(-14) -12
-2(14) 12
4(-7) -3
-4(7) 3
Of all the integer factors of -28 only one pair sums to 3. Hence
x² + 3x - 28 = (x - 4)(x + 7)
Trinomials with leading coefficient some number other than one.
Consider (ax + b)(cx + d) = acx² + adx + bcx + bd.
= acx² + (ad + bc)x + bd.
The trinomial given on the right side of the equation above is harder
to factor than a trimonial with leading coefficient one since you have
to consider all possible factors of the first term of the trinomial as
well as all possible factors of the last term of the trinomial and find
which of these pairs results in the combined middle term. This
procedure is called factoring by trial and error.
Example: Factor 6x² + 11x + 4
Possible factors of 6x² are : 6x(x) and 2x(3x).
Possible factors of 4 are: 4(1), 2(2), -4(-1), and -2(-2). Since
the sign of the middle term is positive we need only consider positive
factors of the last term for negative factors of last term would lead
to a middle term which would be negative.
Try 6x(x) with 4(1): (6x + 4)(x + 1) = 6x² + 6x + 4x + 1 = 6x² + 10x + 4 (Does not work.)
(6x + 1)(x + 4) = 6x² + 24x + x + 4 = 6x² + 25x + 4 (Does not work.)
Try 6x(x) with 2(2): (6x + 2)(x + 2) = 6x² + 12x + 2x + 4 = 6x² + 14x + 4 (Does not work.)
Try 3x(2x) with 4(1) : (3x + 1)(2x + 4) = 6x² + 12x + 2x + 4 = 6x² + 14x + 4 (Does not work.)
(3x + 4)(2x + 1) = 6x² +3x + 8x + 4 = 6x² + 11x + 4 Works!!!
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