Unit 3, lesson 23

Factoring and Recognizing Structure

How to rewrite a polynomial as a product using common factors, trinomial patterns, special products and grouping, and how to confirm any factorization by expanding it back.

  • Theory part 45
  • Practice part 46
  • Powers and Expressions

Theory, part 45

Practice, part 46

Expanding turns a product into a sum. Factoring runs the same road backward, and it is usually the harder direction, because a sum gives no label saying which product it came from. The good news is that the same few patterns show up again and again. Learn to spot them, and every answer can be checked in a few seconds by multiplying it back out.

See also Expanding and Factoring for a Purpose in Foundations for Calculus.

Key terms

Factor

Definition To factor an expression means to write it as an equivalent product of two or more expressions. Each expression in that product is a factor of the original. For polynomials with integer coefficients, factoring usually means writing the polynomial as a product of polynomials of lower degree, again with integer (or at least real) coefficients.

Factoring is a rewriting step. It produces an equivalent expression and does not by itself solve anything. Only when the factored expression is set equal to 00 does it lead to solutions.

In plain words A factored form is the same expression dressed as a multiplication. You can always test it by expanding, and the result must match the original exactly.

Example The factorization x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3) checks out, since (x−3)(x+3)=x2+3x−3x−9=x2−9(x - 3)(x + 3) = x^2 + 3x - 3x - 9 = x^2 - 9. Note that x2−9x^2 - 9 alone is not an equation, so there is nothing to solve yet. The equation x2−9=0x^2 - 9 = 0 is a different matter, and its solutions are 33 and −3-3.

Greatest common factor

Definition The greatest common factor (GCF) of the terms of a polynomial is the product of the greatest common factor of the coefficients and, for each variable, the lowest power of that variable appearing in every term. By the distributive property,

ab+ac=a(b+c),ab + ac = a(b + c),

so the GCF can be pulled out in front, leaving the quotient of each term by the GCF inside the parentheses.

In plain words Find what every term shares, write it once outside, and keep what is left of each term, sign included, inside.

Example In 6x2−10x6x^2 - 10x both terms share 22 and xx, so the GCF is 2x2x. Dividing each term by 2x2x leaves 3x3x and −5-5, so

6x2−10x=2x(3x−5).6x^2 - 10x = 2x(3x - 5).

Expanding gives back 6x2−10x6x^2 - 10x. Pulling out 2x2x is always the first move, because it makes everything after it smaller.

Quadratic trinomial

Definition A quadratic trinomial is a polynomial ax2+bx+cax^2 + bx + c with a≠0a \ne 0, usually with b≠0b \ne 0 and c≠0c \ne 0 so that it has three terms. If a=1a = 1 and there are numbers pp and qq with p+q=bp + q = b and pq=cpq = c, then

x2+bx+c=(x+p)(x+q).x^2 + bx + c = (x + p)(x + q).

If a≠1a \ne 1, look for numbers pp and qq with pq=acpq = ac and p+q=bp + q = b, split the middle term as px+qxpx + qx, and factor by grouping. Not every trinomial factors with integer coefficients.

In plain words For x2+bx+cx^2 + bx + c, hunt for two numbers that multiply to cc and add to bb. Signs matter, so a positive cc with a negative bb means both numbers are negative.

Example For x2−7x+12x^2 - 7x + 12 the numbers −3-3 and −4-4 multiply to 1212 and add to −7-7, so x2−7x+12=(x−3)(x−4)x^2 - 7x + 12 = (x - 3)(x - 4). For 2x2+7x+32x^2 + 7x + 3 the product acac is 66, and the pair 11 and 66 adds to 77,

2x2+7x+3=2x2+x+6x+3=x(2x+1)+3(2x+1)=(2x+1)(x+3).\begin{aligned} &2x^2 + 7x + 3 \\[0.4em] &= 2x^2 + x + 6x + 3 \\[0.4em] &= x(2x + 1) + 3(2x + 1) \\[0.4em] &= (2x + 1)(x + 3). \end{aligned}

Difference of squares

Definition For all real numbers aa and bb,

a2−b2=(a−b)(a+b).a^2 - b^2 = (a - b)(a + b).

A sum of squares behaves differently. If k>0k > 0 and c>0c > 0, then kx2+ckx^2 + c is positive for every real xx, so it has no real zero. A linear factor px+qpx + q with real p≠0p \ne 0 would be zero at x=−qpx = -\frac{q}{p}, so kx2+ckx^2 + c has no linear factor with real coefficients and does not factor over the real numbers.

In plain words Two perfect squares with a minus sign between them always split into a minus version and a plus version. A quadratic made of a positive square plus a positive number, such as x2+4x^2 + 4, does not factor over the real numbers.

Example The expression 9x2−259x^2 - 25 is (3x)2−52(3x)^2 - 5^2, so

9x2−25=(3x−5)(3x+5).9x^2 - 25 = (3x - 5)(3x + 5).

Expanding gives 9x2+15x−15x−25=9x2−259x^2 + 15x - 15x - 25 = 9x^2 - 25, and the middle terms cancel exactly as they should. By contrast, 9x2+259x^2 + 25 has no real factorization into linear factors, and guesses like (3x+5)2(3x + 5)^2 fail because that expands to 9x2+30x+259x^2 + 30x + 25.

Perfect square trinomial

Definition For all real numbers aa and bb,

a2+2ab+b2=(a+b)2,a2−2ab+b2=(a−b)2.\begin{aligned} a^2 + 2ab + b^2 &= (a + b)^2, \\[0.4em] a^2 - 2ab + b^2 &= (a - b)^2. \end{aligned}

A trinomial is a perfect square exactly when its first and last terms are squares a2a^2 and b2b^2 and its middle term is 2ab2ab or −2ab-2ab.

In plain words If the ends are squares and the middle is twice their product, the whole thing is one binomial squared. The sign of the middle term is the sign inside the parentheses.

Example In x2−10x+25x^2 - 10x + 25 the ends are x2x^2 and 525^2, and the middle term is −2⋅x⋅5=−10x-2 \cdot x \cdot 5 = -10x. So x2−10x+25=(x−5)2x^2 - 10x + 25 = (x - 5)^2. Checking, (x−5)2=x2−5x−5x+25=x2−10x+25(x - 5)^2 = x^2 - 5x - 5x + 25 = x^2 - 10x + 25.

Factoring by grouping

Definition Factoring by grouping splits the terms of a polynomial into groups, factors each group so that the same expression BB appears in every group, and then pulls out BB with the distributive property,

A1B+A2B=(A1+A2)B.A_1 B + A_2 B = (A_1 + A_2) B.

It works only when the groups really do share the same factor, which sometimes requires pulling out a negative number from a group.

In plain words Factor in pairs, and if the same parentheses appear in both pairs, that repeated block becomes a factor of the whole.

Example Factor x3+3x2−4x−12x^3 + 3x^2 - 4x - 12. The first pair gives x2(x+3)x^2(x + 3), and pulling −4-4 from the second pair gives −4(x+3)-4(x + 3). So

x3+3x2−4x−12=x2(x+3)−4(x+3)=(x2−4)(x+3)=(x−2)(x+2)(x+3).\begin{aligned} &x^3 + 3x^2 - 4x - 12 \\[0.4em] &= x^2(x + 3) - 4(x + 3) \\[0.4em] &= (x^2 - 4)(x + 3) \\[0.4em] &= (x - 2)(x + 2)(x + 3). \end{aligned}

The last step uses the difference of squares. Pulling out +4+4 instead would leave 4(−x−3)4(-x - 3), which does not match x+3x + 3.

Nested structure

Definition An expression has nested structure when a whole subexpression plays the role of a single variable. If an expression can be written as P(u)P(u), where PP is a polynomial and uu stands for a subexpression such as x+1x + 1 or x2x^2, then any factorization of P(u)P(u) gives a factorization of the original expression after uu is replaced by that subexpression.

In plain words Treat a repeated block as one letter, factor the simpler expression, then put the block back.

Example In (x+1)2−5(x+1)+6(x + 1)^2 - 5(x + 1) + 6, let u=x+1u = x + 1. Then u2−5u+6=(u−2)(u−3)u^2 - 5u + 6 = (u - 2)(u - 3), and putting x+1x + 1 back gives

(x+1−2)(x+1−3)=(x−1)(x−2).\begin{aligned} &(x + 1 - 2)(x + 1 - 3) \\[0.4em] &= (x - 1)(x - 2). \end{aligned}

Expanding the original confirms it, since x2+2x+1−5x−5+6=x2−3x+2=(x−1)(x−2)x^2 + 2x + 1 - 5x - 5 + 6 = x^2 - 3x + 2 = (x - 1)(x - 2). The same idea factors x4−13x2+36x^4 - 13x^2 + 36 with u=x2u = x^2, which gives (x2−4)(x2−9)(x^2 - 4)(x^2 - 9), and two differences of squares finish it as (x−2)(x+2)(x−3)(x+3)(x - 2)(x + 2)(x - 3)(x + 3).

Check yourself

  1. Which of the following is a factor of 2x2−5x−122x^2 - 5x - 12? (A) x+4x + 4 (B) x−3x - 3 (C) 2x+32x + 3 (D) 2x−32x - 3
Show the answer

The answer is (C). The pair 33 and −8-8 multiplies to ac=−24ac = -24 and adds to −5-5, which leads to 2x2−5x−12=(2x+3)(x−4)2x^2 - 5x - 12 = (2x + 3)(x - 4). Expanding confirms it, since 2x2−8x+3x−12=2x2−5x−122x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12. Choices (A) and (D) come from (2x−3)(x+4)=2x2+5x−12(2x - 3)(x + 4) = 2x^2 + 5x - 12, which has the wrong sign on the middle term. Choice (B) would require x=3x = 3 to make the expression zero, but at x=3x = 3 it equals 18−15−12=−918 - 15 - 12 = -9.

  1. Which expression is equivalent to 16x4−8116x^4 - 81? (A) (4x2−9)2(4x^2 - 9)^2 (B) (4x−9)(4x+9)(4x - 9)(4x + 9) (C) (2x−3)(2x+3)(4x2+9)(2x - 3)(2x + 3)(4x^2 + 9) (D) (2x−3)2(4x2+9)(2x - 3)^2(4x^2 + 9)
Show the answer

The answer is (C). As a difference of squares, 16x4−81=(4x2−9)(4x2+9)16x^4 - 81 = (4x^2 - 9)(4x^2 + 9), and the first factor is again a difference of squares, (2x−3)(2x+3)(2x - 3)(2x + 3). The sum 4x2+94x^2 + 9 does not factor over the real numbers. Choice (A) expands to 16x4−72x2+8116x^4 - 72x^2 + 81. Choice (B) is 16x2−8116x^2 - 81, the wrong power. Choice (D) uses (2x−3)2(2x - 3)^2 where (2x−3)(2x+3)(2x - 3)(2x + 3) is needed.

  1. If x2−y2=48x^2 - y^2 = 48 and x+y=8x + y = 8, what is the value of x−yx - y?
Show the answer

The answer is 66. The difference of squares gives (x−y)(x+y)=48(x - y)(x + y) = 48, so 8(x−y)=488(x - y) = 48 and x−y=6x - y = 6. There is no need to find xx and yy, though they are 77 and 11.

How this lesson connects