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

( x − 20 ) 2 \left(x-20\right)^2 ( x − 20 ) 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) ( x − 4 ) ( x + 100 )

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

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

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

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

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

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 )

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

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

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

− y ( y + 1 3 ) 2 -y\left(y+\frac13\right)^2 − y ( y + 3 1 ) 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, $a^{2} - b^{2}$ 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, with terms like $a^{3} \pm b^{3}$ , factor using the SOAP rule into a binomial times a trinomial, ensuring correct sign placement. Understanding these formulas aids in efficiently factoring polynomials by identifying bases, coefficients, and exponents.

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, the polynomial must meet two key criteria: it must consist of exactly two terms, both of which are perfect squares, and these terms must be separated by a subtraction sign.

For example, consider the polynomial $x^2 - 4$. Here, $x^2$ is a perfect square (since $x \times x = x^2$), and \$4$ is also a perfect square (since \(2 \times 2 = 4$). Because the terms are separated by subtraction, this fits the difference of squares pattern. Using the formula, it factors as:

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

It is important to note that this method does not apply to sums of squares, such as \)a^2 + b^2\(, which cannot be factored over the real numbers using this formula.

Applying this to other examples, consider \)16 - 9x^2$. Both $16$ and $9x^2$ are perfect squares since $16 = 4^2$ and $9x^2 = (3x)^2\(. The expression is a difference, 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, not a difference, and thus cannot be factored using this method.

Understanding how to identify and factor the difference of squares is a fundamental skill in algebra that simplifies polynomial expressions and solves equations efficiently. Always verify that the terms are perfect squares and that they are separated by subtraction before applying this factoring technique.

Problem

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

$$\left$(k-7$\right$)$\left$(k+7$\right$)$

$$\left$(k-7$\right$)$\left$(k-7$\right$)$

$k$\left$(k-49$\right$)$

$$\left$(k-49$\right$)$\left$(k+49$\right$)$

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 (x²)² - 4². Applying the difference of squares formula, which states that a² - b² = (a + b)(a - b), we factor it into (x² + 4)(x² - 4). This is the fully factored form of the original binomial.

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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 original expression may not appear as a difference of squares due to the constants involved, factoring out the GCF simplifies the process. In this case, both terms share a GCF of -6. Factoring out -6 leaves the expression inside the parentheses as 25 - x².

The expression 25 - x² is a classic example of a difference of squares, since both 25 and x² are perfect squares. Recall that the difference of squares formula is:

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

Applying this, rewrite 25 - x² as:

\[5^2 - x^2\]

Then factor it as:

\[ (5 + x)(5 - x) \]

Finally, include the factored-out GCF to express the fully factored form of the original binomial:

\[ -6(5 + x)(5 - x) \]

This approach highlights the importance of first extracting the greatest common factor to reveal underlying structures such as the difference of squares, enabling complete factorization of the binomial.

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, so factor out 3 first. This 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), factor the expression as 3(2y + 3x)(2y - 3x). This method highlights the importance of factoring out the GCF before recognizing special products like the difference of squares, ensuring the binomial is factored completely and correctly.

concept

Perfect Square Trinomials

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Perfect Square Trinomials Video Summary

Factoring perfect square trinomials is a valuable skill 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 determine if a trinomial is a perfect square, first verify that the first and last terms are perfect squares. For example, in the trinomial x² + 10x + 25, the terms x² and 25 are perfect squares since x² = (x)² and 25 = 5². Next, check 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, for the trinomial 4x² − 36x + 81, the first term 4x² is (2x)² and the last term 81 is 9². The middle term −36x equals −2 × 2x × 9, confirming it fits the pattern of 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 a plus or minus sign between the binomial terms. Mastering this approach allows for efficient factoring of perfect square trinomials, which is essential for simplifying expressions and solving quadratic equations.

Problem

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

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

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

$$\left$(x-20$\right$)^2$

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

Problem

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

$open paren 4 m plus 5 close paren squared$

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

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

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

Problem

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

$12$\left$(x+2$\right$)^2$

$12$\left$(x+4$\right$)^2$

$12x$\left$(x+2$\right$)^2$

$12x$\left$(x+4$\right$)^2$

Problem

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

$4y$\left$(y+6$\right$)$\left$(y+1$\right$)$

$4y$\left$(y+3$\right$)^2$

$36y$\left$(y+1$\right$)^2$

$-4y$\left$(y+3$\right$)^2$

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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 it as a perfect square trinomial despite involving two variables. First, identify if the first and last terms are perfect squares. The term x² is a perfect square since it equals x × x. Similarly, 9y² is a perfect square because it equals (3y) × (3y). Next, verify the middle term, 6xy, matches the form 2ab, where a = x and b = 3y. Calculating 2 × x × 3y gives 6xy, confirming the pattern.

Since the middle term is positive, the trinomial factors as the square of a binomial: $(x + 3y)^2$. This method applies the concept of perfect square trinomials, which are expressions of the form $a^2 + 2ab + b^2 = (a + b)^2$. Recognizing these patterns simplifies factoring complex expressions involving multiple variables.

concept

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 consisting of two terms that are perfect cubes, either added or subtracted. Recognizing these forms allows us to factor them efficiently using specific formulas derived from their product patterns. For example, expressions like x³ + 8 or 27y³ - 1 fit this category because both terms are perfect cubes.

The factoring rules for sums and differences of cubes are as follows: a sum of cubes, a³ + b³, factors into (a + b)(a² - ab + b²), while a difference of cubes, a³ - b³, factors into (a - b)(a² + ab + b²). These formulas break down the original cubic expression into a product of a binomial and a trinomial, simplifying the polynomial.

To apply these rules, first identify the cube roots of each term. For instance, in x³ + 8, the cube root of x³ is x, and the cube root of 8 is 2, since 2 × 2 × 2 = 8. Using the sum of cubes formula, this factors to (x + 2)(x² - 2x + 4). Similarly, for 27y³ - 1, recognize that 27 is 3³ and 1 is 1³, so the expression becomes (3y)³ - 1³. Applying the difference of cubes formula yields (3y - 1)(9y² + 3y + 1).

To keep track of the signs when factoring sums or differences of cubes, the acronym SOAP is useful: Same, Opposite, Always Positive. This means the sign in the binomial factor matches the original sign between the cubes (same), the sign of the middle term in the trinomial is the opposite, and the last term in the trinomial is always positive. This helps avoid common mistakes in sign placement during factoring.

Understanding how to factor sums and differences of cubes is essential for simplifying complex polynomials and solving higher-degree equations. By identifying perfect cubes and applying the appropriate formula with the SOAP method, factoring becomes a systematic and manageable process.

Problem

Factor completely.
$27 x cubed plus 125$

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

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

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

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

Problem

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

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

$$\left$(y-$\frac$13$\right$)$\left$(y^2+$\frac$13y+$\frac$19$\right$)$

$y$\left$(y-$\frac$13$\right$)^2$

$-y$\left$(y+$\frac$13$\right$)^2$

example

Sum & Difference of Cubes Example 5

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Sum & Difference of Cubes Example 5 Video Summary

To factor the polynomial $$\frac{2}{3}$ x^3 - $\frac{2}{3}$ y^3$ , start by identifying the greatest common factor (GCF). Both terms share a factor of $$\frac{2}{3}$$ , so factor this out first. This simplifies the expression to $$\frac{2}{3}$ (x^3 - y^3)$ . The remaining 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, $a = x$ and $b = y$ , so the factorization becomes:

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

This method highlights the importance of first extracting the GCF to simplify the polynomial before applying the difference of cubes formula. Recognizing the structure of cubic terms and correctly applying the factorization formula ensures accurate and efficient simplification of expressions involving cubes.

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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 coefficients 16 and 54 share a GCF of 2. Factoring out 2 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:

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

Simplifying the terms inside the second parentheses:

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

Therefore, the fully factored form of the original polynomial is:

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

This process highlights the importance of first extracting the greatest common factor before applying special factoring formulas such as the sum of cubes. Recognizing perfect cubes and correctly applying the sum of cubes identity enables efficient polynomial factorization.

Here’s what students ask on this topic:

The difference of squares is a special polynomial form where you have two terms that are perfect squares separated by a subtraction sign. It looks like $a^{2} - b^{2}$ . To factor it, you use the formula $(a + b)(a - b)$ . This works because when you multiply these two binomials, the middle terms cancel out, leaving the difference of squares. For example, $x^2 - 4$ factors as $(x + 2)(x - 2)$ since $4 = 2^2$ . Remember, this method only applies when both terms are perfect squares and the sign between them is subtraction.

Perfect square trinomials are polynomials of the form $a^2 $\pm$ 2ab + b^2$ . They come from squaring a binomial, such as $(a + b)^2$ or $(a - b)^2$ . To factor a perfect square trinomial, first check if the first and last terms are perfect squares and if the middle term equals $2ab$ . If so, you can factor it as $(a + b)^2$ when the middle term is positive, or $(a - b)^2$ when it is negative. For example, $x^2 + 10x + 25$ factors as $(x + 5)^2$ because $10x = 2 $\times$ x $\times$ 5$ and both $x^2$ and $25$ are perfect squares.

The sum or difference of cubes involves polynomials like $a^3 + b^3$ or $a^3 - b^3$ , where both terms are perfect cubes. These can be factored using special formulas. For the sum of cubes, the factorization is $(a + b)(a^2 - ab + b^2)$ . For the difference of cubes, it is $(a - b)(a^2 + ab + b^2)$ . A helpful acronym to remember the signs is SOAP: Same sign in the binomial as the original expression, Opposite sign in the middle term of the trinomial, and Always Positive for the last term. For example, $x^3 + 8$ factors as $(x + 2)(x^2 - 2x + 4)$ .

The sum of squares, such as $a^2 + b^2$ , cannot be factored over the real numbers using simple binomials like the difference of squares. This is because the sum of squares does not factor into real linear factors due to the positive sign between the terms. Unlike the difference of squares, which can be expressed as $(a + b)(a - b)$ , the sum of squares requires complex numbers for factorization, such as $(a + bi)(a - bi)$ . Therefore, in basic algebra, the sum of squares is considered prime and cannot be factored further.

To factor special polynomials like difference of squares, perfect square trinomials, or sum/difference of cubes, you first identify the terms $a$ and $b$ by recognizing perfect squares or cubes. For example, in $x^2 - 9$ , $a = x$ and $b = 3$ because $9 = 3^2$ . In a perfect square trinomial like $4x^2 + 12x + 9$ , $a = 2x$ and $b = 3$ since $4x^2 = (2x)^2$ and $9 = 3^2$ . For cubes, in $27y^3 - 1$ , $a = 3y$ and $b = 1$ because $27y^3 = (3y)^3$ and $1 = 1^3$ . Identifying these values correctly is essential for applying the factoring formulas.

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Factoring Special Products: Videos & Practice Problems