Integrated Math 2–3 Honors · Lesson Review · September 28, 2026

Factoring Differences of Squares

Spot two perfect-square terms separated by subtraction, write conjugate factors, and keep factoring until every factor is prime.

13-minute edited lesson9 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Recognize a Missing Linear Term 1:1902Irrational Factors Still Work 3:1603Sum or Difference of Squares?4:1804The Difference-of-Squares Identity 5:4805Perfect-Square Coefficients 7:1306Two Variables and Irrational Squares 8:2507Factor Higher Powers Completely 9:5708Repeat the Pattern 11:4709Greatest Common Factor First

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02 · Understand

Learning targets

Recognize $A^2-B^2$ even when the square roots are variables or irrational numbers.
Factor a difference of squares as conjugate binomials.
Distinguish a difference of squares from a sum of squares over the real numbers.
Remove a GCF first and repeat the pattern for higher powers.

03 · Connect

Key ideas

Every expression uses KaTeX with semantic MathML for visual clarity and assistive technology.

01Missing linear term: $x^2-49=x^2+0x-49$ Treat the missing term as a coefficient of zero.
02Core identity: $a^2-b^2=(a-b)(a+b)$ The factors have the same terms with opposite signs.
03Irrational factors: $x^2-5=(x-\sqrt5)(x+\sqrt5)$ Perfect-square terms may have irrational square roots and still factor over the reals.
04Sum versus difference: $x^2+36\text{ does not factor over }\mathbb{R}$ The real-number pattern requires subtraction.
05Factor completely: $40x^5-250x^3=10x^3(2x-5)(2x+5)$ Take out the GCF, apply difference of squares, and check every remaining factor.

04 · Apply

Worked examples

These are the structures developed in the recorded lesson.

Basic pattern

Find the square root of each term

x^2-49=(x-7)(x+7)

The roots are $x$ and $7$ , so the factors are conjugates.

Coefficients

Recognize a square monomial

100x^2-81=(10x-9)(10x+9)

Because $100x^2=(10x)^2$ and $81=9^2$ , use $A=10x$ and $B=9$ .

Two variables

The pattern is unchanged

16x^2-25y^2=(4x-5y)(4x+5y)

Both terms are perfect-square monomials.

Higher degree

Check the factors again

x^4-16=(x-2)(x+2)(x^2+4)

After the first split, $x^2-4$ is another difference of squares, while $x^2+4$ is prime over the reals.

Repeated pattern

Continue until every factor is prime

x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)

05 · Avoid

Common mistakes

Factoring a sum as conjugates.

$A^2+B^2$ does not use the real-number difference-of-squares pattern.

Forgetting square roots.

The factors use $A$ and $B$ , not the original squared terms.

Stopping too early.

Recheck each factor for another difference of squares.

Skipping the GCF.

Remove the greatest common factor before using a special pattern.

06 · Check yourself

Try it first

These problems use new structures—not repeated worked examples. Factor completely before opening each solution.

1Factor

18x^5-72x^3

. What must happen before the special pattern?

$18x^5-72x^3=18x^3(x-2)(x+2)$ Remove the GCF first; the remaining binomial is a difference of squares.

2Factor

81r^4-1

completely.

$81r^4-1=(3r-1)(3r+1)(9r^2+1)$ Apply the pattern twice. The sum $9r^2+1$ is prime over the reals.

3Error analysis: is

m^2+81\stackrel{?}{=}(m-9)(m+9)

true?

$(m-9)(m+9)=m^2-81\ne m^2+81$ Conjugates produce a difference, not a sum.

4Factor the two-variable expression

9a^4-16b^6

$9a^4-16b^6=(3a^2-4b^3)(3a^2+4b^3)$ The square roots are $3a^2$ and $4b^3$ .

5Challenge: factor

y^8-256

completely.

$y^8-256=(y-2)(y+2)(y^2+4)(y^4+16)$ The first difference of squares creates another one; the two sums remain prime over the reals.

07 · Revisit

Transcript and captions

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