Learn on PengienVision, Algebra 2Chapter 3: Polynomial Functions

Lesson 3: Polynomial Identities

In this Grade 11 enVision Algebra 2 lesson, students learn to prove and apply polynomial identities including the Difference of Squares, Square of a Sum, Sum of Cubes, and Difference of Cubes to multiply, factor, and simplify expressions. Students also explore Pascal's Triangle and the Binomial Theorem to expand powers of binomials such as (x + y)^n. The lesson builds fluency in recognizing perfect squares and perfect cubes within polynomial expressions and selecting the appropriate identity to rewrite them efficiently.

Section 1

Binomial Squares Pattern

Property

If aa and bb are real numbers,

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

This pattern is a shortcut for squaring a binomial.
To use it, you square the first term, square the last term, and then add or subtract double the product of the two terms.

Examples

  • To multiply (z+6)2(z+6)^2, we use the pattern (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. Here, a=za=z and b=6b=6. The result is (z)2+2(z)(6)+(6)2(z)^2 + 2(z)(6) + (6)^2, which simplifies to z2+12z+36z^2 + 12z + 36.
  • To multiply (3y+4)2(3y+4)^2, we identify a=3ya=3y and b=4b=4. Using the pattern, we get (3y)2+2(3y)(4)+(4)2(3y)^2 + 2(3y)(4) + (4)^2. This simplifies to 9y2+24y+169y^2 + 24y + 16.

Section 2

Factoring a Difference of Squares

Property

A difference of squares is a perfect square subtracted from a perfect square. It can be rewritten as two factors containing the same terms but opposite signs.

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

To factor, confirm both terms are perfect squares and write the factored form. A sum of squares cannot be factored.

Examples

  • To factor 49x2−1649x^2 - 16, recognize this as (7x)2−42(7x)^2 - 4^2. The factors are (7x+4)(7x−4)(7x+4)(7x-4).
  • To factor 100y4−81z2100y^4 - 81z^2, recognize this as (10y2)2−(9z)2(10y^2)^2 - (9z)^2. The factors are (10y2+9z)(10y2−9z)(10y^2+9z)(10y^2-9z).
  • To factor a2−1a^2 - 1, recognize this as a2−12a^2 - 1^2. The factors are (a+1)(a−1)(a+1)(a-1).

Explanation

When you see a perfect square minus another perfect square, it factors into two identical binomials, one with a plus and one with a minus. This causes the middle terms from FOIL to cancel out, leaving just the first and last terms.

Section 3

Cube of a Binomial

Property

  1. (x+y)3=x3+3x2y+3xy2+y3(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
  2. (x−y)3=x3−3x2y+3xy2−y3(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

Examples

  • To expand (a+4)3(a + 4)^3, use the first formula with x=ax=a and y=4y=4: a3+3(a)2(4)+3(a)(4)2+43=a3+12a2+48a+64a^3 + 3(a)^2(4) + 3(a)(4)^2 + 4^3 = a^3 + 12a^2 + 48a + 64.
  • To expand (2b−1)3(2b - 1)^3, use the second formula with x=2bx=2b and y=1y=1: (2b)3−3(2b)2(1)+3(2b)(1)2−13=8b3−12b2+6b−1(2b)^3 - 3(2b)^2(1) + 3(2b)(1)^2 - 1^3 = 8b^3 - 12b^2 + 6b - 1.
  • To expand (z2+3)3(z^2 + 3)^3, use the first formula with x=z2x=z^2 and y=3y=3: (z2)3+3(z2)2(3)+3(z2)(3)2+33=z6+9z4+27z2+27(z^2)^3 + 3(z^2)^2(3) + 3(z^2)(3)^2 + 3^3 = z^6 + 9z^4 + 27z^2 + 27.

Explanation

These formulas are shortcuts for expanding expressions like (a+b)3(a+b)^3 without performing a lengthy multiplication. Memorizing these patterns for the sum and difference of two cubed terms saves time and helps prevent common algebra mistakes.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Binomial Squares Pattern

Property

If aa and bb are real numbers,

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

This pattern is a shortcut for squaring a binomial.
To use it, you square the first term, square the last term, and then add or subtract double the product of the two terms.

Examples

  • To multiply (z+6)2(z+6)^2, we use the pattern (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. Here, a=za=z and b=6b=6. The result is (z)2+2(z)(6)+(6)2(z)^2 + 2(z)(6) + (6)^2, which simplifies to z2+12z+36z^2 + 12z + 36.
  • To multiply (3y+4)2(3y+4)^2, we identify a=3ya=3y and b=4b=4. Using the pattern, we get (3y)2+2(3y)(4)+(4)2(3y)^2 + 2(3y)(4) + (4)^2. This simplifies to 9y2+24y+169y^2 + 24y + 16.

Section 2

Factoring a Difference of Squares

Property

A difference of squares is a perfect square subtracted from a perfect square. It can be rewritten as two factors containing the same terms but opposite signs.

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

To factor, confirm both terms are perfect squares and write the factored form. A sum of squares cannot be factored.

Examples

  • To factor 49x2−1649x^2 - 16, recognize this as (7x)2−42(7x)^2 - 4^2. The factors are (7x+4)(7x−4)(7x+4)(7x-4).
  • To factor 100y4−81z2100y^4 - 81z^2, recognize this as (10y2)2−(9z)2(10y^2)^2 - (9z)^2. The factors are (10y2+9z)(10y2−9z)(10y^2+9z)(10y^2-9z).
  • To factor a2−1a^2 - 1, recognize this as a2−12a^2 - 1^2. The factors are (a+1)(a−1)(a+1)(a-1).

Explanation

When you see a perfect square minus another perfect square, it factors into two identical binomials, one with a plus and one with a minus. This causes the middle terms from FOIL to cancel out, leaving just the first and last terms.

Section 3

Cube of a Binomial

Property

  1. (x+y)3=x3+3x2y+3xy2+y3(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
  2. (x−y)3=x3−3x2y+3xy2−y3(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

Examples

  • To expand (a+4)3(a + 4)^3, use the first formula with x=ax=a and y=4y=4: a3+3(a)2(4)+3(a)(4)2+43=a3+12a2+48a+64a^3 + 3(a)^2(4) + 3(a)(4)^2 + 4^3 = a^3 + 12a^2 + 48a + 64.
  • To expand (2b−1)3(2b - 1)^3, use the second formula with x=2bx=2b and y=1y=1: (2b)3−3(2b)2(1)+3(2b)(1)2−13=8b3−12b2+6b−1(2b)^3 - 3(2b)^2(1) + 3(2b)(1)^2 - 1^3 = 8b^3 - 12b^2 + 6b - 1.
  • To expand (z2+3)3(z^2 + 3)^3, use the first formula with x=z2x=z^2 and y=3y=3: (z2)3+3(z2)2(3)+3(z2)(3)2+33=z6+9z4+27z2+27(z^2)^3 + 3(z^2)^2(3) + 3(z^2)(3)^2 + 3^3 = z^6 + 9z^4 + 27z^2 + 27.

Explanation

These formulas are shortcuts for expanding expressions like (a+b)3(a+b)^3 without performing a lengthy multiplication. Memorizing these patterns for the sum and difference of two cubed terms saves time and helps prevent common algebra mistakes.