Factor completely. x 2 − 40 x + 400 x^2-40x+400

( x − 20 ) 2 \left(x-20\right)^2

( x − 20 ) ( x + 20 ) \left(x-20\right)\left(x+20\right)

( x − 4 ) ( x − 100 ) \left(x-4\right)\left(x-100\right)

( x − 4 ) ( x + 100 ) \left(x-4\right)\left(x+100\right)

Factor completely. 4 m 2 + 20 m + 25 4m^2+20m+25

( 2 m + 5 ) 2 \left(2m+5\right)^2

( 4 m + 5 ) 2 \left(4m+5\right)^2

( 2 m + 5 ) ( m + 1 ) \left(2m+5\right)\left(m+1\right)

( 4 m + 5 ) ( m + 5 ) \left(4m+5\right)\left(m+5\right)

Factor completely. 27 x 3 + 125 27x^3+125

( 3 x + 5 ) ( 9 x 2 − 15 x + 25 ) \left(3x+5\right)\left(9x^2-15x+25\right)

( 3 x − 5 ) ( 9 x 2 + 15 x + 25 ) \left(3x-5\right)\left(9x^2+15x+25\right)

( 27 x + 5 ) ( x 2 + 25 ) \left(27x+5\right)\left(x^2+25\right)

( 3 x + 5 ) ( x 2 + 5 ) \left(3x+5\right)\left(x^2+5\right)

Factor completely. y 3 − 1 27 y^3-\frac{1}{27}

( y − 1 3 ) ( y 2 + 1 3 y + 1 9 ) \left(y-\frac13\right)\left(y^2+\frac13y+\frac19\right)

( y + 1 3 ) ( y 2 + 1 3 y + 1 9 ) \left(y+\frac13\right)\left(y^2+\frac13y+\frac19\right)

y ( y − 1 3 ) 2 y\left(y-\frac13\right)^2

− y ( y + 1 3 ) 2 -y\left(y+\frac13\right)^2

7. Factoring

Factoring Special Products

7. Factoring

Factoring Special Products: Videos & Practice Problems

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Topic summary

Factoring special polynomials involves recognizing patterns like the difference of squares, perfect square trinomials, and sums or differences of cubes. For a difference of squares, a2 - b2 factors as (a + b)(a - b). Perfect square trinomials, where the first and last terms are perfect squares and the middle term equals 2ab, factor as (a \pm b)^2. Sums or differences of cubes, like a^3 \pm b^3, factor using the SOAP rule into a binomial times a trinomial, ensuring correct sign placement. Understanding these patterns aids in efficiently factoring polynomials.

concept

Difference of Squares

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Difference of Squares Video Summary

Factoring special polynomials often involves recognizing common patterns, one of the most important being the difference of squares. This occurs when an expression is the subtraction of two perfect square terms. The difference of squares can be factored using the formula:

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

This formula is derived from the product of conjugates, where multiplying the sum and difference of the same two terms results in the difference of their squares. To apply this factoring method, first confirm that both terms are perfect squares and that they are separated by a subtraction sign.

For example, consider the expression $x^2 - 4$. Here, $x^2$ is a perfect square since it is $x \times x$, and 4 is a perfect square since it is $2 \times 2$. Because the terms are subtracted, this fits the difference of squares pattern. Using the formula, it factors as:

\[x^2 - 4 = (x + 2)(x - 2)\]

It is crucial to note that this method only applies to differences of squares. Expressions like $a^2 + b^2$, which are sums of squares, cannot be factored using this formula.

Consider the expression $16 - 9x^2$. Both 16 and $9x^2$ are perfect squares since $16 = 4^2$ and $9x^2 = (3x)^2$. The subtraction sign confirms it is a difference of squares, so it factors as:

\[16 - 9x^2 = (4 + 3x)(4 - 3x)\]

On the other hand, an expression like $y^2 + 25$ is a sum of squares, since both terms are perfect squares but are added rather than subtracted. This expression cannot be factored using the difference of squares formula.

Understanding how to identify and factor differences of squares is a foundational skill in algebra that simplifies polynomial expressions and solves equations efficiently. Always verify the presence of two perfect squares and a subtraction sign before applying this factoring technique.

Problem

Write the following difference of squares as a product of two binomials.
$k squared minus 49$

$(k - 7) (k + 7)$

$(k - 7) (k - 7)$

$k (k - 49)$

$(k - 49) (k + 49)$

Problem

Write the following difference of squares as a product of two binomials.
$81 minus G squared$

$(81 - G) (81 + G)$

$(9 - G^{2}) (9 + G^{2})$

$(9 - G) (9 - G)$

$(9 - G) (9 + G)$

example

Difference of Squares Example 1

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Difference of Squares Example 1 Video Summary

To factor the binomial completely, recognize that it represents a difference of squares. Although it may not appear obvious initially, the term x⁴ can be expressed as (x²)², and 16 is a perfect square since it equals 4². This allows us to rewrite the expression as:

\[(x^2)^2 - 4^2\]

Applying the difference of squares formula, which states that a² - b² = (a + b)(a - b), we factor the expression into:

\[(x^2 + 4)(x^2 - 4)\]

This is the fully factored form of the original binomial. Note that while the first factor, x² + 4, cannot be factored further over the real numbers, the second factor, x² - 4, is itself a difference of squares and can be factored further if needed.

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Difference of Squares Example 2

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Difference of Squares Example 2 Video Summary

To factor the binomial completely, start by identifying the greatest common factor (GCF). Although the terms may not initially appear to form a difference of squares, factoring out the GCF can simplify the expression. In this case, both terms share a GCF of -6. Factoring out -6 gives:

$-6(25 - x^2)$

Inside the parentheses, the expression 25 - x² is a difference of squares because 25 is \$5^2$ and $x^2\( is a perfect square. The difference of squares formula is:

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

Applying this to \)25 - x^2\(, we get:

\)25 - x^2 = (5 + x)(5 - x)\(

Therefore, the completely factored form of the original binomial is:

\)-6(5 + x)(5 - x)$

This method highlights the importance of factoring out the greatest common factor first to reveal recognizable patterns such as the difference of squares, enabling complete factorization.

example

Difference of Squares Example 3

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Difference of Squares Example 3 Video Summary

To factor the binomial 12y² - 27x² completely, start by identifying the greatest common factor (GCF) of the coefficients. The numbers 12 and 27 share a GCF of 3. Factoring out 3 simplifies the expression to 3(4y² - 9x²). The remaining binomial inside the parentheses is a difference of squares because both 4y² and 9x² are perfect squares. Specifically, 4y² = (2y)² and 9x² = (3x)². Applying the difference of squares formula, which states that a² - b² = (a + b)(a - b), the expression factors to 3(2y + 3x)(2y - 3x). This method highlights the importance of first factoring out the GCF before recognizing and applying special factoring formulas like the difference of squares.

concept

Perfect Square Trinomials

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Perfect Square Trinomials Video Summary

Factoring perfect square trinomials is a key skill in algebra that builds on understanding special polynomial forms. A perfect square trinomial arises when a binomial is squared, resulting in three terms: the first and last terms are perfect squares, and the middle term is twice the product of the square roots of these terms. Specifically, the expressions a² + 2ab + b² and a² − 2ab + b² factor into (a + b)² and (a − b)² respectively.

To factor a perfect square trinomial, first identify if the first and last terms are perfect squares. For example, in the trinomial x² + 10x + 25, x² is the square of x, and 25 is the square of 5. Next, verify if the middle term equals 2ab, where a and b are the square roots of the first and last terms. Here, 10x equals 2 × x × 5, confirming it is a perfect square trinomial. Since the middle term is positive, the factorization is (x + 5)².

Similarly, consider the trinomial 4x² − 36x + 81. The first term, 4x², is the square of 2x, and the last term, 81, is the square of 9. The middle term, −36x, matches −2 × 2x × 9, confirming it as a perfect square trinomial with a negative middle term. Therefore, it factors as (2x − 9)².

Recognizing the sign of the middle term is crucial because it determines whether the factorization uses addition or subtraction inside the squared binomial. This method streamlines factoring by reversing the process of squaring binomials, making it easier to simplify expressions and solve equations involving perfect square trinomials.

Problem

Factor completely.
$x^{2} - 40 x + 400$

$(x - 20) (x + 20)$

$(x - 4) (x - 100)$

$open paren x minus 20 close paren squared$

$(x - 4) (x + 100)$

Problem

Factor completely.
$4 m^{2} + 20 m + 25$

$open paren 4 m plus 5 close paren squared$

$open paren 2 m plus 5 close paren squared$

$(2 m + 5) (m + 1)$

$(4 m + 5) (m + 5)$

Problem

Factor completely. Hint: Factor out the GCF first.
$12 x^{2} + 48 x + 48$

$12 {(x + 2)}^{2}$

$12 {(x + 4)}^{2}$

$12 x {(x + 2)}^{2}$

$12 x {(x + 4)}^{2}$

Problem

Factor completely. Hint: Factor out the GCF first.
$- 4 y^{3} - 24 y^{2} - 36 y$

$4 y (y + 6) (y + 1)$

$4 y {(y + 3)}^{2}$

$36 y {(y + 1)}^{2}$

$- 4 y {(y + 3)}^{2}$

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Perfect Square Trinomials Example 4

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Perfect Square Trinomials Example 4 Video Summary

To factor the trinomial x² + 6xy + 9y² completely, recognize that it is a perfect square trinomial involving two variables, x and y. First, identify if the first and last terms are perfect squares. The term x² is a perfect square since it can be expressed as x × x. Similarly, 9y² is a perfect square because it equals (3y) × (3y).

Next, verify the middle term, 6xy, to see if it fits the form of 2ab, where a and b are the square roots of the first and last terms respectively. Here, a = x and b = 3y. Calculating 2ab gives:

\[2ab = 2 \times x \times 3y = 6xy\]

Since this matches the middle term, the trinomial is indeed a perfect square. Therefore, it factors as:

\[x^2 + 6xy + 9y^2 = (x + 3y)^2\]

This method highlights the importance of recognizing perfect square trinomials and applying the formula for factoring them, even when multiple variables are involved.

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Sum & Difference of Cubes

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Sum & Difference of Cubes Video Summary

A sum or difference of cubes refers to a polynomial expression where two terms are perfect cubes being added or subtracted, such as $ x^3 + 8 $. Recognizing these expressions allows us to factor them using specific formulas derived from their product patterns. When factoring a sum of cubes, the expression $ a^3 + b^3 $ can be factored into the product of a binomial and a trinomial: $ (a + b)(a^2 - ab + b^2) $. Similarly, a difference of cubes, $ a^3 - b^3 $, factors as $ (a - b)(a^2 + ab + b^2) $.

To apply these formulas, first identify the cube roots of each term. For example, in $ x^3 + 8 $, $ x^3 $ is a perfect cube with root $ x $, and $ 8 $ is a perfect cube with root $ 2 $. Using the sum of cubes formula, this factors to $ (x + 2)(x^2 - 2x + 4) $. The key to correctly factoring these expressions lies in managing the signs within the factors, which can be remembered using the acronym SOAP: Same, Opposite, Always Positive. This means the sign in the binomial matches the original sign between the cubes, the sign before the middle term in the trinomial is the opposite, and the last term in the trinomial is always positive.

For a difference of cubes example, consider $ 27y^3 - 1 $. Recognize that $ 27y^3 = (3y)^3 $ and $ 1 = 1^3 $. Applying the difference of cubes formula, it factors as $ (3y - 1)(9y^2 + 3y + 1) $. Here, the binomial uses the same minus sign, the trinomial’s middle term uses the opposite plus sign, and the last term remains positive, consistent with the SOAP rule.

Understanding how to factor sums and differences of cubes is essential for simplifying polynomials and solving equations efficiently. By identifying perfect cubes and applying the appropriate formulas with careful attention to signs, factoring these special polynomials becomes a straightforward process that enhances algebraic manipulation skills.

Problem

Factor completely.
$27 x cubed plus 125$

$(3 x - 5) (9 x^{2} + 15 x + 25)$

$(3 x + 5) (9 x^{2} - 15 x + 25)$

$(27 x + 5) (x^{2} + 25)$

$(3 x + 5) (x^{2} + 5)$

Problem

Factor completely.
$y^{3} - \frac{1}{27}$

$(y + \frac{1}{3}) (y^{2} + \frac{1}{3} y + \frac{1}{9})$

$(y - \frac{1}{3}) (y^{2} + \frac{1}{3} y + \frac{1}{9})$

$y {(y - \frac{1}{3})}^{2}$

$- y {(y + \frac{1}{3})}^{2}$

example

Sum & Difference of Cubes Example 5

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Sum & Difference of Cubes Example 5 Video Summary

To factor the polynomial two thirds x cubed minus two thirds y cubed, start by recognizing that both terms contain a common factor of two thirds. Although the expression involves cubes, the coefficient two thirds is not a perfect cube, so it cannot be factored as a cube root directly. Instead, factor out the greatest common factor (GCF), which is two thirds, from the entire expression.

Factoring out the GCF, the polynomial becomes:

\[\frac{2}{3}(x^3 - y^3)\]

Now, the expression inside the parentheses is a difference of cubes, which can be factored using the formula for the difference of cubes:

\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]

Here, a corresponds to x and b corresponds to y. Applying the formula, the factorization is:

\[\frac{2}{3}(x - y)(x^2 + xy + y^2)\]

This factorization breaks down the original polynomial into a product of a constant, a binomial, and a trinomial, which is useful for simplifying expressions or solving equations involving this polynomial. Understanding how to factor out the GCF first and then apply the difference of cubes formula is essential for efficiently handling polynomials with coefficients that are not perfect cubes but share common factors.

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Sum & Difference of Cubes Example 6

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Sum & Difference of Cubes Example 6 Video Summary

To factor the polynomial $16x^3 + 54y^3$, start by identifying the greatest common factor (GCF). The number 16 factors into $2 \times 8$, and 8 further factors into $2 \times 4$, with 4 breaking down into $2 \times 2$. The number 54 factors into $2 \times 27$, and 27 breaks down into $3 \times 9$, with 9 further factoring into $3 \times 3$. The only common factor between 16 and 54 is 2, so the GCF is 2.

Factoring out the GCF of 2 from the polynomial gives:

\[2(8x^3 + 27y^3)\]

Now, observe that $8x^3$ and $27y^3$ are perfect cubes since $8 = 2^3$ and $27 = 3^3$. This allows us to recognize the expression inside the parentheses as a sum of cubes, which can be factored using the sum of cubes formula:

\[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]

Here, $a = 2x$ and $b = 3y$. Applying the formula, we get:

\[2 \times \left( (2x + 3y) \left( (2x)^2 - (2x)(3y) + (3y)^2 \right) \right)\]

Calculating each term inside the second parentheses:

\[(2x)^2 = 4x^2, \quad (2x)(3y) = 6xy, \quad (3y)^2 = 9y^2\]

Substituting these back, the fully factored form of the polynomial is:

\[2(2x + 3y)(4x^2 - 6xy + 9y^2)\]

This factorization simplifies the original polynomial by extracting the greatest common factor and applying the sum of cubes formula, which is essential for breaking down expressions involving perfect cubes into products of binomials and trinomials.

Here’s what students ask on this topic:

The formula for factoring a difference of squares is based on the pattern where you have two perfect square terms separated by a subtraction sign. If you have an expression of the form $a^{2} - b^{2}$ , it factors into the product of two binomials: $(a + b) (a - b)$ . This works only when both terms are perfect squares and the sign between them is negative. For example, $x^{2} - 4$ factors as $(x + 2) (x - 2)$ because $x^{2}$ and $2^{2}$ are perfect squares. Remember, this method does not apply to sums of squares like $a^{2} + b^{2}$ , which cannot be factored using this formula.

Perfect square trinomials are polynomials of the form $a^{2} + 2 a b + b^{2}$ or $a^{2} - 2 a b + b^{2}$ . To factor them, first confirm that the first and last terms are perfect squares and that the middle term equals $2ab$ or $-2ab$ . If the middle term is positive, the trinomial factors as $(a + b)^{2}$ . If the middle term is negative, it factors as $(a - b)^{2}$ . For example, $x^{2} + 10 x + 25$ factors as $(x + 5)^{2}$ because $x^2$ and $25$ are perfect squares and $10x = 2 imes x imes 5$ . This method simplifies factoring trinomials that fit this pattern.

The SOAP method is a helpful acronym to remember the sign pattern when factoring sums or differences of cubes. For an expression like $a^{3} + b^{3}$ (sum of cubes) or $a^{3} - b^{3}$ (difference of cubes), the factorization is a binomial times a trinomial. SOAP stands for Same, Opposite, Always Positive: the sign in the binomial is the same as the original sign between the cubes; the first sign in the trinomial is opposite; and the last sign in the trinomial is always positive. Specifically, $a^{3} + b^{3}$ factors as $(a + b) (a^{2} - a b + b^{2})$ , and $a^{3} - b^{3}$ factors as $(a - b) (a^{2} + a b + b^{2})$ . This method ensures correct sign placement in the factors.

No, you cannot factor a sum of squares using the difference of squares formula. The difference of squares formula applies only when you have two perfect square terms separated by a subtraction sign, like $a^{2} - b^{2}$ . A sum of squares, such as $a^{2} + b^{2}$ , does not factor over the real numbers using simple factoring methods. It is considered prime in basic algebra. More advanced techniques involving complex numbers can factor sums of squares, but in beginning and intermediate algebra, sums of squares are left unfactored. For example, $x^2 + 25$ cannot be factored using the difference of squares formula because both terms are positive, and the formula requires a subtraction sign.

To identify if a polynomial is a perfect square trinomial, check three conditions: first, the polynomial must have three terms; second, the first and last terms must be perfect squares; third, the middle term must be equal to twice the product of the square roots of the first and last terms. Mathematically, if the polynomial is $a^{2} + 2 a b + b^{2}$ or $a^{2} - 2 a b + b^{2}$ , it is a perfect square trinomial. For example, in $x^2 + 10x + 25$ , $x^2$ and $25$ are perfect squares, and the middle term $10x$ equals $2 imes x imes 5$ . This confirms it is a perfect square trinomial and factors as $(x + 5)^2$ .

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Factoring Special Products: Videos & Practice Problems