XVII Factoring

Factoring is the opposite of distributing. A polynomial expression is transformed into a monomial by using various strategies that depend on the type of expression that is to be factored.

Greatest Common (shared) Factor (GCF)

The GCF of a polynomial can be identified by finding all factors of each of its terms and finding the highest factors common to all terms. Example:

12x212x^2: {1,2,3,4,6,12,x,x2}{1,2,3,4,6,12,x,x^2}
15x415x^4: {1,3,5,15,x,x2,x3,x4}{1,3,5,15,x,x^2,x^3,x^4 }

The GCF here is 3x23x^2

A negative coefficient has the same factors as its positive equivalent! 9x2,15x4,63x-9x^2,15x^4, 6 \rarr 3x

A greatest common factor may consist only of variables if the coefficients lack any common factors: 11ab3,a2b3,ab2ab211ab^3, a^2b^3,ab^2 \rarr ab^2

Factoring using the GCF

e.g. 6x+146x + 14

  1. Identify the GCF 2\rarr 2
  2. Divide each term by the GCF 3x+7\rarr 3x + 7
  3. Multiply the resulting polynomial by the GCF 2(3x+7)\rarr 2(3x+7)

When the coefficients are fractions whose denominators are the same, the simplify into their numerators:

59x5+19x429x3=19x3(5x2+x2)\frac 5 9 x^5 + \frac 1 9 x^4 - \frac 2 9 x^3 = \frac 1 9 x^3(5x^2 + x -2) (GCF: 19x3\frac 1 9 x^3

Factoring by Group

Given that 8x+7x=x(8+7)8x + 7x = x(8+7), shared factors can be grouped up and factored with a term of their coefficients.

7(x+3)+y(x+3)=(x+3)(7+y)7(x+3) + y(x+3) = (x+3)(7+y)

9xy3(a+b)(a+b)=(a+b)(9xy31)9xy^3(a+b) -(a+b) = (a+b)(9xy^3-1)

The GCF of a set of terms can be used to create groups:

xy+2x+3y+6=x(y+2)+3(y+2)=(y+2)(x+3)xy + 2x + 3y + 6 = x(y+2) + 3(y+2) = (y+2)(x+3)

It's important for signs to carry over in step 1:

2xy+5y24x10y=y(2x+5y)2(2x+5y)=(2x+5y)(y2)2xy + 5y^2 - 4x - 10y = y(2x+5y) -2(2x +5y) = (2x+5y)(y-2)

Terms can be switched around if their pairs otherwise lack any common factors:

3xy4+x12y=3xy+x412y3xy-4+x-12y = 3xy+x -4-12y

The first step is to divide by the GCF if all terms of a polynomial thay may be factored by group has common factors:

15xy+15yz5xz5z215xy + 15yz - 5xz - 5z^2 (GCF: 5)
=5[3xy+3yzxzz2]=5[3y(x+z)z(x+z)]=5(x+z)(3yz)= 5[3xy+3yz-xz-z^2] = 5[3y(x+z)-z(x+z)] = 5(x+z)(3y-z)

Factoring Trinomials

Assuming the worst case where there are no common factors but 1.


The way to solve these is by drawing a diamond diagram. That is an ×\times with bb in the north position and acac in the south position. The numbers west and east have to be found. They ought to:

  1. Sum up to bb
  2. Factor up to acac

If a=1a = 1 a shortcut can be used to skip directly to the factored form of the original expression. In this case the numbers found are to be grouped with the variable. For example, assuming dd and ee to be numbers found in the diagram, the solution would simply be (x+d)(x+e)(x+d)(x+e).

To factor any polynomial one has to ask in order:

  1. Is the expression in order? (Terms with higher exponents first)
  2. Is there a GCF?
  3. Are thre 4 terms? 3 Terms?
  4. If 3 terms, does a=1a = 1?

A trinomial can not be factored if using the diamond diagram does not yield any two numbers.

The power of the variable in the factors depends on the highest power of the expression:
x4+10x2+21=(x2+3)(x2+7)x^4+10x^2 + 21 = (x^2+3)(x^2+7)

If an expression's aa equals to anything but 11, the numbers found using the diamond diagram are used to construct two terms that replace the original expressions bb term:

5x525x430x35x35x^5-25x^4-30x^3 \rarr 5x^3
=5x3[x2+5x6]d=6,e=1= 5x^3[x^2+5x-6] \rarr d=-6, e = 1
5x3[x2+x6x6]=5x3[x(x+1)6(x+1)]5x^3[x^2+x-6x-6] = 5x^3[x(x+1)-6(x+1)]
=5x3(x+1)(x6)= 5x^3(x+1)(x-6)

If aa is a negative the first step is to factor all terms by 1-1:

5x219x+4=1(5x2+19x4)-5x^2-19x+4 = -1(5x^2+19x-4) (important: change signs of terms).

There's another shortcut for factoring by group

  1. Divide numbers found in diagram by aa
  2. Write as fractions (simplify!)
  3. Construct terms by pairing denominator ×x\times x and adding or subtracting their numerators

e.g. given d=3d = -3 e=8e = 8 and a=2a = 2

32,41(82)(2x3)(x+4)\frac {-3}{2}, \frac 4 1 (\frac 8 2) \rarr (2x-3)(x+4)

Factoring Binomials

The diamong diagram method may be used in some cases of binomials by realizing that b=0b = 0. However in most cases it yields no results for dd and ee. e.g. x29x^2-9:

=(x3)(x+3)= (x-3)(x+3)

However a case such as x215x^2-15 yields no possible values for cc and ee. For some cases of binomials there are patterns that can be matched to perform to factor them.

Difference of Squares

a2b2=(ab)(a+b)a^2 - b^2 = (a-b)(a+b)


x216=x242=(x+4)(x4)x^2 - 16 = x^2 - 4^2 = (x+4)(x-4)

If one term is a fraction and both numberator and denominator possess perfect square roots, the fraction can be simplified to their roots and the fraction squared.

x21625=x2(45)2=(x45)(x+45)x^2-\frac {16}{25} = x^2 - (\frac 4 5)^2 = (x-\frac 4 5)(x+\frac 4 5)

Remember 12=11^2 = 1

16y21=(4y)2(1)2=(4y1)(4y+1)16y^2 - 1 = (4y)^2 - (1)^2 = (4y-1)(4y+1)

Exponents of higher powers can be forced into squares

x4y14=(x2)2(y7)2=(x2y7)(x2+7y)x^4 - y^{14} = (x^2)^2 - (y^7)^2 = (x^2-y^7)(x^2+7y)

One resulting factor may by a difference of squares in itself

x416=(x2)242=(x24)(x2+4)x^4 - 16 = (x^2)^2 - 4^2 = (x^2-4)(x^2+4)

... x24x^2-4 is another difference of squares therefore further factoring into (x2)(x+2)(x2+4)(x-2)(x+2)(x^2+4) is required.

If applicable the binomial is to be factored using the GFC first

9y325y=y(9y225)=y(3y5)(3y+5)9y^3-25y = y(9y^2-25) = y(3y-5)(3y+5)

As with trinomials negative factors of the highest power are to be factored out first

25x2+121=(25x2121)=(5x11)(5x+11)-25x^2 + 121 = -(25x^2 - 121) = -(5x-11)(5x+11)

Difference of Cubes

a3b3=(ab)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2+ab+b^2)

Sum of Cubes

a3+b3=(a+b)(a2abb2)a^3+b^3 = (a+b)(a^2-ab_b^2)

These usually result in factores that can not be factored any further.

Nothing can be done if there are four terms that can not be grouped

Quadratic Equations

Equations with a highest power of 2.

Standard Form: ax2+bx+c=0ax^2 + bx + c = 0

The first step in dealing with these is always to put them into standard form!

Zero Product Rule
a×b=0a \times b = 0 implies that either aa or bb is guaranteed to be 00.

e.g. (x+1)(x3)=0(x+1)(x-3)=0 implies that either of the following has to be true:

  1. x+1=0x+1=0
  2. x3=0x-3=0

Therefore the equation has two solutions for xx, namely x=1x=-1 and x=3x=3.