Factoring Polynomials by Types

FACTORING POLYNOMIALS BY TYPE

TYPE

I GCF (greatest common factor)….. no matter how complicated, always do this first!

ab + ac = a(b + c)

ab + a = a(b + 1)

II Difference of two perfect squares:

a2 – b2 = (a – b)(a + b)

36x2 – 49 = (6x – 7)(6x + 7) use FOIL to check: 36x2 + 42x – 42x – 49

III Perfect square trinomials:

The marks of a Type III: First and last terms are perfect squares, middle term is an even number, and the last term is always positive!

(note – on all types of trinomials, the middle term is always key when you multiply the outside terms and add them to the product of the inside terms; always test the middle term, and if it checks out, you most likely have it properly factored!)

a2 + 2ab + b2 = (a + b)2

a2 – 2ab + b2 = (a – b)2

16x2 + 24x + 9 = (4x + 3)2

To check: since the middle term equals 2ab when you FOIL it, multiply 4x (first term of the factor) times 3 (last term of the factor) and then times 2 = 24x; the middle term checks!

25a2 – 40ab + 16b2 = (5a – 4b)2

To check: multiply 5a times - 4b, then times 2 = -40ab, the same as the middle term, which checks. The sign of the middle term of the perfect square trinomial always determines the sign of the factor.

IV Imperfect square trinomial :

x2 + bx + c (find the factors of c such that their sum equals b)

for help with signs see chart:

b c factored signs example

+ + = ( + )( + ) x2 + 9x + 20 = (x + 4)(x + 5)

– – = ( – largest number)( + ) x2 – x – 20 = (x – 5)(x + 4)

+ – = ( +largest number)( – ) x2 + x – 20 = (x + 5)(x – 4)

– + = ( – )( – ) x2 – 9x + 20 = (x – 5)(x – 4)

Again, always check the middle term by adding the products of the inside terms and the outside terms: example 1: 4x + 5x = 9x; example 2: -5x + 4x = -x; example 3: 5x – 4x = x; and example 4: -5x – 4x = -9x

V Imperfect square trinomial with first term coefficient:

ax2 + bx + c (like a Type IV, but you must factor both a and c, such that their sum using FOIL equals b)

3x2 + 17x + 24 = (3x + 8)(x + 3)

4x2 – 4x – 15 = (2x – 5)(2x + 3)

On Type V’s you must check all the factors of the first and last terms and all their combinations of sums such that when you add their products you again get the middle term.

VI 4-term polynomial

Factor using the following strategies:

1) separate it into 2 binomials such that when you take out the GCF of each, you have a common binomial factor, e.g. ax + a + b + bx

First, rearrange it into pairs with like factors: ax + bx + a + b, recognizing that you have

a + b twice as a factor: x(a + b) + 1(a + b), then factor out the now common factor a + b, so that you now have (a + b)(x + 1)

2) separate into the difference of a perfect square trinomial (Type III) and a perfect square: a2 + 2ab + b2 – x2

Now factor the trinomial: (a + b)2 – x2, and now you can see that you have a difference of 2 perfect squares, which factors into (a + b – x)(a + b + x)

another example: x2 – a2 – 2ab – b2 = x2 – (a2 + 2ab + b2) = x2 – (a + b)2, then

[x – (a + b)][x + (a + b)], or (x – a – b)(x + a + b)

VII Perfect cube binomial:

a3 + b3 = (a + b)(a2 – ab + b2)

a3 – b3 = (a – b)(a2 + ab + b2)