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, Polynomials, and Polynomial Functions

Special Products

6. Exponents, Polynomials, and Polynomial Functions

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 enhances efficiency beyond FOIL, especially with multivariable polynomials and numerical coefficients, reinforcing mastery of polynomial degrees and terms.

Concept

Squaring Binomials

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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 in this form is:

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

This means the square of the binomial equals 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 later on.

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

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

Here, the sign of the middle term matches the sign in the binomial, while the last term remains positive. For (3x - 1)², treat 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 such as factoring, quadratic equations, and polynomial identities. Mastery of these special product formulas enhances problem-solving speed and accuracy in algebra.

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Follow along with each video using our printable worksheets

Problem

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

$9 y squared plus 25$

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

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

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

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

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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: a² − b². This formula is consistent regardless of the order of the terms, meaning a + b times a − b or a − b times a + b will always yield 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 expression x + 7 multiplied by x − 7. Here, a is x and b is 7. Applying the difference of squares formula, the product becomes:

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

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

More complex expressions can also be handled using this formula. For instance, with 5x − 3 times 5x + 3, identify a = 5x and b = 3. Squaring these gives:

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

Again, expanding with FOIL would show the middle terms canceling, confirming the difference of squares result. This formula is a powerful tool for simplifying products of conjugates efficiently, especially when dealing with polynomials or algebraic expressions.

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

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Multiply Using the Difference of Squares Formula Example 1 Video Summary

When multiplying binomials, recognizing special products can simplify the process and help identify patterns in the results. For example, squaring a binomial such as 5x + 7 involves the perfect square trinomial formula, which is expressed as (a + b)² = a² + 2ab + b². Here, a is 5x and b is 7. Squaring a gives 25x² because both the coefficient and the variable are squared, while squaring b results in 49. The middle term is twice the product of a and b, calculated as 2 × 5x × 7 = 70x. Thus, the expanded form is 25x² + 70x + 49, a perfect square trinomial.

When the binomial is squared but with a subtraction, such as (5x - 7)², the formula remains the same but the middle term changes sign. This yields a² - 2ab + b², resulting in 25x² - 70x + 49. The only difference from the previous case is the negative middle term, reflecting the subtraction in the binomial.

Multiplying a sum and difference of the same two terms, like (5x + 7)(5x - 7), uses the difference of squares formula: (a + b)(a - b) = a² - b². This product eliminates the middle term entirely, leaving 25x² - 49. This highlights how the sum and difference multiplication results in a simpler expression without the linear term.

Comparing these results reveals consistent elements: each expression begins with 25x² and ends with 49. The presence and sign of the middle term distinguish the perfect square trinomials from the difference of squares. Understanding these patterns aids in quickly expanding binomials and recognizing the structure of polynomial expressions, which is fundamental in algebraic manipulation and problem-solving.

Here's what students ask on this topic:

The formula for squaring a binomial of the form $a^{2} + 2ab + b^{2}$ applies when you have an expression like (a + b)². To use 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 expressions like (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 factors and is especially useful for simplifying expressions quickly without 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 $25x^{2} - 6x + 1$ after applying the formula. Recognizing perfect square trinomials helps in factoring and simplifying polynomial expressions efficiently.

When squaring a binomial with complex terms, such as (3x - 1)², treat each term as a whole. For example, here a = 3x and b = 1. Square the entire first term: $(3x)^{2} = 9x^{2}$ , then calculate twice the product: $2 * 3x * 1 = 6x$ , and finally square the second term: $1^{2} = 1$ . The expanded form is $9x^{2} - 6x + 1$ . Remember to distribute exponents to all parts of the term, not just the variable, to avoid common mistakes.

Conjugates are pairs of binomials that differ only in the sign between their two terms, such as (a + b) and (a - b). Multiplying conjugates results in the difference of squares formula: $a^{2} - b^{2}$ . For example, (5x + 3)(5x - 3) expands to $25x^{2} - 9$ . Recognizing conjugates allows you to quickly simplify products without expanding each term individually, which is especially useful in algebraic manipulations.