Multiply the binomials. ( y + 1 8 ) ( y − 1 8 ) \left(y+\frac18\right)\left(y-\frac18\right)

y 2 − 1 4 y − 1 64 y^2-\frac14y-\frac{1}{64}

y 2 + 1 4 y − 1 64 y^2+\frac14y-\frac{1}{64}

6. Exponents and Polynomials

Special Products

6. Exponents and Polynomials

Special Products: Videos & Practice Problems

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 + b)^2, producing perfect square trinomials. For (a - b)^2, the middle term is subtracted. Multiplying conjugates, (a + b)(a - b), uses the difference of squares formula a2 - b2, eliminating the middle terms. Recognizing these patterns enhances efficiency in polynomial operations involving terms, coefficients, and exponents.

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, often faster than applying the FOIL method manually. 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 expands into three terms: 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 trinomial is known as a perfect square trinomial, a key concept in algebra that often appears in factoring and solving quadratic equations.

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

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

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 these formulas allows for quicker expansion of binomials and recognition of perfect square trinomials, which are foundational in algebraic manipulation and solving quadratic expressions efficiently.

Problem

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

$9 y squared plus 25$

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

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

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

Problem

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

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

$x^{4} + 1$

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

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

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. This formula simplifies the multiplication process by producing an expression with only two terms: the square of the first term minus the square of the second term. Specifically, the product is given by the equation:

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

This formula is especially useful because it always results in a difference (subtraction) between the squares of the two terms, regardless of the order of the factors. The terms a + b and a − b are known as conjugates, and multiplying conjugates always yields a difference of squares.

For example, consider the expression x + 7 multiplied by x − 7. Here, a = x and b = 7. Applying the difference of squares formula, the product is:

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

This approach is often faster and more straightforward than using the FOIL method, as the middle terms cancel out automatically.

In cases where a and b are more complex expressions, the formula still applies. For instance, with the product 5x − 3 times 5x + 3, identify a = 5x and b = 3. Squaring these gives:

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

Remember to square both the coefficient and the variable when squaring terms like 5x. This method efficiently simplifies multiplication of conjugates 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.
$(3 m + 4 n) (3 m - 4 n)$

$3 n squared minus 4 n squared$

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

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

$9 m squared minus 16 n squared$

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

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

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

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

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

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 using the formula for the square of a sum: (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, which is a perfect square trinomial.

When the binomial is squared but with a subtraction, such as (5x - 7)², the formula (a - b)² = a² - 2ab + b² applies. The terms a² and b² remain the same, but the middle term changes sign, becoming negative. This results in 25x² - 70x + 49. This subtle difference highlights how the sign between terms affects the middle term in the expansion.

Multiplying a sum and difference of the same terms, like (5x + 7)(5x - 7), uses the difference of squares formula: (a + b)(a - b) = a² - b². Here, the middle terms cancel out, leaving only the difference between the squares of a and b. This simplifies to 25x² - 49, with no middle term present.

Comparing these results reveals consistent elements: each expression contains the terms 25x² and 49, representing the squares of the original binomial components. The key variation lies in the middle term, which appears as either positive, negative, or absent, depending on whether the binomial is squared with addition, squared with subtraction, or multiplied as a sum and difference. Understanding these patterns aids in quickly identifying and expanding binomial products, reinforcing the importance of special product formulas in algebra.

Here’s what students ask on this topic:

The formula for squaring a binomial of the form $a^{2} + 2ab + b^{2}$ applies to expressions like $(a, +, b) 2$ . It expands to $a^{2} + 2ab + b^{2}$ . To use it, identify the terms $a$ and $b$ in your binomial. Square the first term, add twice the product of the two terms, then add the square of the second term. For example, for $(y, +, 5) 2$ , $a = y$ and $b = 5$ . The expansion is $y^{2} + 2(y)(5) + 25 = y^2 + 10y + 25$ . This method is faster than FOIL and helps avoid missing the middle term.

When multiplying expressions like $(a, +, b) \cdot (a, -, b)$ , you use the difference of squares formula: $a^{2} - b^{2}$ . This product eliminates the middle terms, resulting in just two terms. For example, $(x + 7)(x - 7) = x^2 - 49$ . Here, $a = x$ and $b = 7$ . This formula works regardless of the order of the factors and is a quick alternative to FOIL, especially when dealing with conjugates (sum and difference pairs).

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, squaring $(3x - 1)^2$ gives $9x^2 - 6x + 1$ , which is a perfect square trinomial. Recognizing these helps simplify polynomial expressions quickly without expanding fully using FOIL.

To square a binomial like $(3, x, -, 1) 2$ , treat the entire first term as $a = 3x$ and the second term as $b = 1$ . Square the first term by squaring both the coefficient and the variable: $(3x)^2 = 9x^2$ . Then calculate the middle term: $-2 imes 3x imes 1 = -6x$ . Finally, square the second term: $1^2 = 1$ . Putting it all together, the expansion is $9x^2 - 6x + 1$ . This method ensures you correctly apply exponents to coefficients and variables.

Conjugates are pairs of binomials that have the same terms but opposite signs, such as $(a + b)$ and $(a - b)$ . Multiplying conjugates uses the difference of squares formula: $(a + b)(a - b) = a^2 - b^2$ . This product eliminates the middle terms, simplifying the expression quickly. Recognizing conjugates helps in factoring and simplifying expressions efficiently, especially when dealing with polynomials that can be rewritten as differences of squares.