Square each $$binomial.(x2+1)2(x^2$+1)^2$$

$$x4+2x2+1x^4$+2x^2+1x4+2x2+1$$

$$x2+2x2+1x^2$+2x^2+1x2+2x2+1$$

Multiply the binomials.(y+18)(y−18)\left(y+\frac18\right)\left(y-\frac18\right)

y2−$$164y^2$$-\frac{1}{64}y2−641

y2−14y−$$164y^2$$-\frac14y-\frac{1}{64}

y2+14y−$$164y^2$$+\frac14y-\frac{1}{64}

$$y2+164y^2$$+\frac{1}{64}y2+641

Multiply the binomials.(3m+4n)(3m−4n)(3m+4n)(3m-4n)

9m2−$$16n29m^2$-16n^29m2$$−16n2

Multiply the binomials.(p3q+7p2)(p3q−$$7p2)(p^3q+7p^2$)(p^3$q-7p^2) $$

p6q2−$$49p4p^6q^2$-49p^4p6q2$$−49p4

$$p6q2+49p4p^6q^2$+49p^4p6q2+49p4$$

p6q2−14p5q−$$49p4p^6q^2$-14p^5$q-49p^4p6q2$$−14p5q−49p4

p6q2+14p5q−$$49p4p^6q^2$+14p^5$q-49p^4p6q2+14p5q$$−49p4

6. Exponents and Polynomials

Special Products

6. Exponents and Polynomials

Special Products: Videos & Practice Problems

Video Lessons Worksheet

Topic summary

Understanding how to multiply binomials using special product formulas simplifies working with polynomials. Squaring a binomial follows the formula a2+2ab+b2 for a+b2, producing perfect square trinomials. The difference of squares formula, a2-b2, applies when multiplying conjugates like a+b and a-b. Recognizing these patterns aids in efficiently handling polynomials, exponents, and terms with numerical coefficients and variables.

Concept

Squaring Binomials

Video duration:

Squaring Binomials Video Summary

Squaring a binomial is a common algebraic operation that can be simplified using special product formulas, making the process faster and less error-prone than expanding with the FOIL method. When you have an expression like y + 5 squared, it fits the pattern of (a + b)², where a and b represent any terms—letters, numbers, or combinations thereof. The formula for squaring a binomial is:

\[ (a + b)^2 = a^2 + 2ab + b^2 \]

This means the square of the first term plus twice the product of the two terms plus the square of the second term. A common mistake is to think that (a + b)² equals a² + b², but the middle term 2ab is essential and cannot be omitted.

For example, in (y + 5)², identify a = y and b = 5. Applying the formula:

\[ y^2 + 2 \times y \times 5 + 5^2 = y^2 + 10y + 25 \]

This result is known as a perfect square trinomial, a quadratic expression that arises from squaring a binomial. Recognizing perfect square trinomials is useful for factoring and solving quadratic equations.

When the binomial involves subtraction, such as (3x - 1)², the formula adjusts slightly to:

\[ (a - b)^2 = a^2 - 2ab + b^2 \]

The sign of the middle term matches the sign in the binomial, while the last term is always positive. In this case, a = 3x and b = 1. Remember to square the entire term 3x, not just the variable:

\[ (3x)^2 = 9x^2 \]

Applying the formula:

\[ 9x^2 - 2 \times 3x \times 1 + 1^2 = 9x^2 - 6x + 1 \]

Understanding how to square binomials efficiently helps simplify algebraic expressions and prepares you for more advanced topics like factoring, quadratic equations, and polynomial identities. Mastery of these formulas enhances problem-solving speed and accuracy in algebra.

Study Smarter with Worksheets.

Follow along with each video using our printable worksheets

Problem

Square each binomial.
$(- 3 y + 5)^{2}$

$9 y squared plus 25$

$9y^2-30y+25$

$-9y^2+30y-25$

$-9y^2-30y+25$

Problem

Square each binomial.
$open paren x squared plus 1 close paren squared$

$x^4+2x^2+1$

$x^4+1$

$x^2+2x^2+1$

$x^2+2x+1$

Concept

Multiply Using the Difference of Squares Formula

Video duration:

Multiply Using the Difference of Squares Formula Video Summary

When multiplying the product of the sum and difference of the same two terms, such as a + b times a − b, the result follows the difference of squares formula. Unlike squaring a binomial, which produces a perfect square trinomial, this product simplifies to just two terms: the square of the first term minus the square of the second term. This can be expressed as:

\[ (a + b)(a - b) = a^2 - b^2 \]

This formula always holds true regardless of the order of the factors, meaning a − b times a + b yields the same result. The terms a + b and a − b are known as conjugates, a key concept when working with these expressions.

For example, consider the product x + 7 times x − 7. Here, a is x and b is 7. Applying the difference of squares formula, the product simplifies to:

\[ x^2 - 7^2 = x^2 - 49 \]

This method is often faster than using the FOIL method because the middle terms cancel out, leaving only the difference of the squares.

The formula also applies when a and b are more complex expressions. For instance, in the product 5x − 3 times 5x + 3, a is 5x and b is 3. Squaring these gives:

\[ (5x)^2 - 3^2 = 25x^2 - 9 \]

Again, expanding this using FOIL would show the middle terms canceling, confirming the difference of squares result. Understanding and applying this formula simplifies multiplication of conjugate binomials and is a valuable tool in algebraic manipulation.

Problem

Multiply the binomials.
$(y + \frac{1}{8}) (y - \frac{1}{8})$

$y^{2} - \frac{1}{4} y - \frac{1}{64}$

$y^{2} + \frac{1}{4} y - \frac{1}{64}$

$y^2-$\frac{1}{64}$$

$y^2+$\frac{1}{64}$$

Problem

Multiply the binomials.
$(3 m + 4 n) (3 m - 4 n)$

$3 n squared minus 4 n squared$

$9 m^{2} + 24 m n - 16 n^{2}$

$9 m^{2} - 24 m n - 16 n^{2}$

$9m^2-16n^2$

Problem

Multiply the binomials.
$(x^{2} - 1) (x^{2} + 1)$

$x squared minus 1$

$x^{4} + 2 x^{2} - 1$

$x^{4} - 2 x^{2} + 1$

$x^{4} - 1$

Problem

Multiply the binomials.
$(p^{3} q + 7 p^{2}) (p^{3} q - 7 p^{2})$

$p^6q^2-49p^4$

$p^6q^2+49p^4$

$p^6q^2-14p^5q-49p^4$

$p^6q^2+14p^5q-49p^4$

Example

Multiply Using the Difference of Squares Formula Example 1

Video duration:

Multiply Using the Difference of Squares Formula Example 1 Video Summary

When multiplying binomials, recognizing special products can simplify the process significantly. For example, squaring a binomial such as $ (5x + 7)^2 $ results in a perfect square trinomial. This follows the formula $ (a + b)^2 = a^2 + 2ab + b^2 $, where $ a = 5x $ and $ b = 7 $. Squaring $ a $ gives $ 25x^2 $ because both the coefficient and the variable are squared, and squaring $ b $ yields 49. The middle term is twice the product of $ a $ and $ b $, calculated as $ 2 \times 5x \times 7 = 70x $. Thus, the expanded form is:

\[(5x + 7)^2 = 25x^2 + 70x + 49\]

Similarly, when squaring a binomial with a subtraction, such as $ (5x - 7)^2 $, the formula $ (a - b)^2 = a^2 - 2ab + b^2 $ applies. The only difference from the previous case is the sign of the middle term, which becomes negative:

\[(5x - 7)^2 = 25x^2 - 70x + 49\]

In contrast, multiplying a sum and difference of the same two terms, like $ (5x + 7)(5x - 7) $, results in a difference of squares. This product follows the identity $ (a + b)(a - b) = a^2 - b^2 $, which eliminates the middle term entirely. Applying this to our terms gives:

\[(5x + 7)(5x - 7) = 25x^2 - 49\]

Comparing these results reveals common patterns: all expressions contain the terms $ 25x^2 $ and 49, but the presence and sign of the middle term distinguish the perfect square trinomials from the difference of squares. Understanding these relationships helps in quickly identifying and expanding binomial products, enhancing algebraic manipulation skills and problem-solving efficiency.

Here's what students ask on this topic:

The formula for squaring a binomial of the form $a^{2} + 2ab + b^{2}$ is used when you have an expression like (a + b)². To apply it, identify the terms a and b in your binomial. Then, square the first term (a²), add twice the product of the two terms (2ab), and finally add the square of the second term (b²). For example, for (y + 5)², a = y and b = 5, so the expanded form is $y^{2} + 2* y * 5 + 5^{2}$ , which simplifies to $y^{2} + 10y + 25$ . This method is faster and less error-prone than using FOIL for squaring binomials.

When multiplying the product of the sum and difference of the same two terms, such as (a + b)(a - b), you use the difference of squares formula: $a^{2} - b^{2}$ . This means you square the first term, subtract the square of the second term, and the middle terms cancel out. For example, (x + 7)(x - 7) expands to $x^{2} - 7^{2}$ , which simplifies to $x^{2} - 49$ . This formula works regardless of the order of the binomials and is a quick alternative to FOIL.

A perfect square trinomial is the result of squaring a binomial and has the form $a^{2} + 2ab + b^{2}$ or $a^{2} - 2ab + b^{2}$ . It consists of three terms: the square of the first term, twice the product of the two terms, and the square of the second term. For example, (3x - 1)² expands to $9x 2 - 6x + 1$ , which is a perfect square trinomial. Recognizing these trinomials helps simplify factoring and polynomial operations.

When squaring a binomial with coefficients and variables, such as (3x - 1)², treat the entire term as a single unit. For example, a = 3x and b = 1. Square the first term by squaring both the coefficient and the variable: $(3x)^2 = 9x^2$ . Then calculate twice the product: $2 * 3x * 1 = 6x$ . Finally, square the second term: $1^2 = 1$ . Putting it all together with the correct signs gives $9x^2 - 6x + 1$ . This method ensures you correctly handle exponents and coefficients.

Conjugates are pairs of binomials that have the same terms but opposite signs, such as (a + b) and (a - b). When you multiply conjugates, the result is a difference of squares: $a^{2} - b^{2}$ . This happens because the middle terms cancel out. For example, (5x + 3)(5x - 3) equals $25x^2 - 9$ . Recognizing conjugates helps simplify multiplication of binomials quickly and efficiently.