Special Forms, and Expressions Quadratic in Form
A.SSE.2
Read the standard text
A.SSE.2 Use the structure of an expression to identify ways to rewrite it. For example, to factor 3x(x − 5) + 2(x − 5), students should recognize that the "x − 5" is common to both expressions being added, so it simplifies to (3x + 2)(x − 5); or see x^4 − y^4 as (x^2)^2 − (y^2)^2, thus recognizing it as a difference of squares that can be factored as (x^2 − y^2)(x^2 + y^2).
Section 1 · LearnLearn the idea
In this lesson you will: Factor a difference of squares, a^2-b^2=(a+b)(a-b); Recognize a perfect square trinomial and factor it in one step; Know that a sum of two squares does not factor over the real numbers; Use structure to factor an expression that is quadratic in form, such as x^4-y^4.
A difference of squares splits on sight
x^2-36 has no middle term, and both pieces are perfect squares. Expressions like that factor instantly: (x+6)(x-6). Multiply it back and the middle terms +6x and -6x cancel, which is why the middle term was missing in the first place.
The leading coefficient may be anything here, as long as it is a perfect square: 9x^2-25=(3x+5)(3x-5), and 16x^2-1=(4x+1)(4x-1). Ohio's limits are looser for special forms than for general trinomials, precisely because you recognize these rather than search for them.
Perfect square trinomials
(x+7)^2 expands to x^2+14x+49. Read that backwards: the outer terms are squares, and the middle term is twice the product of their square roots, because the x and the 7 meet twice in the multiplication.
So x^2+14x+49=(x+7)^2 and 4x^2-20x+25=(2x-5)^2. In the second, sqrt(4x^2)=2x, sqrt(25)=5, and 2*2x*5=20x, which matches the middle term in size. The minus sign in the middle puts the minus inside the parentheses.
Quadratic in form
x^4-y^4 is not a quadratic. But x^4=(x^2)^2 and y^4=(y^2)^2, so it behaves like one: a difference of two squares whose square roots happen to be x^2 and y^2.
That gives (x^2+y^2)(x^2-y^2), and the second factor is itself a difference of squares, so the complete factorization is (x^2+y^2)(x+y)(x-y). The first factor is a sum of squares and stops there.
The same trick handles x^4-13x^2+36: think of it as (x^2)^2-13(x^2)+36, factor as (x^2-4)(x^2-9), then split both.
Common mistake — Factoring a sum of squares
Asked to factor x^2+16, a student writes (x+4)^2, or sometimes (x+4)(x-4).
The pull is understandable. Both pieces are perfect squares, the difference of squares rule is fresh, and the expression looks like it ought to cooperate.
It does not. (x+4)^2=x^2+8x+16, which has a middle term the original never had. (x+4)(x-4)=x^2-16, with the wrong sign on the constant. A sum of two squares cannot be factored into binomials with real coefficients at all — there is no pair of real numbers whose product is +16 and whose sum is 0.
How to catch it: Substitute a number. At x=1, x^2+16=17, while (x+4)^2=25 and (x+4)(x-4)=-15. Neither matches, so neither is a factorization. When two terms are added and both are squares, the answer is usually that it does not factor.
Section 2 · DoPractice this standard
These published practice sets are scored on the student site and cover A.SSE.2 directly.