Factoring Special Products: Videos & Practice Problems

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 polynomial \(x^2 - 4\). 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\( represent a sum of squares and cannot be factored 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, as both terms are positive perfect squares, but the sign is addition. This cannot be factored using the difference of squares method.

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 terms are perfect squares and that the operation between them is subtraction before applying the formula.

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 \(a^2 - b^2 = (a + b)(a - b)\), we factor the expression into:

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

This is the fully factored form of the original binomial. Note that the second factor, \(x^2 - 4\), is itself a difference of squares and can be further factored if needed, but as per the problem's scope, the factorization stops here.

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 leaves the expression inside the parentheses as 25 - x².

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

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

Here, 25 can be written as 5² and x² remains as is, so the expression becomes:

\[5^2 - x^2\]

Applying the difference of squares rule, this factors into:

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

Finally, reintroduce the factored-out GCF -6 to write the fully factored form of the original binomial as:

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

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

To factor the expression (z + 5)² - 81 completely, recognize that it fits the pattern of a difference of squares. The term (z + 5)² is a perfect square because it represents (z + 5)(z + 5), and 81 is also a perfect square since it equals 9². This allows us to rewrite the expression as (z + 5)² - 9². Applying the difference of squares formula, which states that a² - b² = (a + b)(a - b), we factor it into (z + 5 + 9)(z + 5 - 9). Simplifying the terms inside the parentheses results in (z + 14)(z - 4). This is the fully factored form of the original binomial, demonstrating how recognizing perfect squares and applying the difference of squares rule can simplify expressions efficiently.

To factor the expression (z + 5)² - 81 completely, recognize that it fits the difference of squares pattern. The term (z + 5)² is a perfect square binomial, representing (z + 5)(z + 5), and 81 is also a perfect square since it equals 9². This allows us to rewrite the expression as (z + 5)² - 9². Applying the difference of squares formula, which states that a² - b² = (a + b)(a - b), we factor it into (z + 5 + 9)(z + 5 - 9). Simplifying the terms inside the parentheses results in (z + 14)(z - 4). This factorization breaks down the original binomial into the product of two linear binomials, providing a complete and simplified factorization.

Factoring perfect square trinomials is a key skill in algebra that builds on the concept of squaring binomials. When you square a binomial such as \(a + b\) or \(a - b\), the result is a perfect square trinomial. Specifically, the expansion of \((a + b)^2\) is \(a^2 + 2ab + b^2\), and the expansion of \((a - b)^2\) is \(a^2 - 2ab + b^2\). Recognizing these patterns allows you to factor trinomials by working backwards.

To factor a perfect square trinomial, first verify that the first and last terms are perfect squares. For example, in the trinomial \(x^2 + 10x + 25\), the first term \(x^2\) is the square of \(x\), and the last term \(25\) is the square of \(5\). Next, check if the middle term equals \(2ab\), where \(a\) and \(b\) are the square roots of the first and last terms respectively. Here, \(10x\) equals \(2 \times x \times 5\), confirming it fits the pattern. Since the middle term is positive, the trinomial factors as \((x + 5)^2\).

Similarly, for the trinomial \(4x^2 - 36x + 81\), the first term \(4x^2\) is \((2x)^2\), and the last term \(81\) is \(9^2\). The middle term \(-36x\) equals \(-2 \times 2x \times 9\), matching the pattern with a negative sign. This means the trinomial factors as \((2x - 9)^2\).

In summary, perfect square trinomials have the form \(a^2 \pm 2ab + b^2\), and factoring them involves identifying \(a\) and \(b\) such that the middle term is twice their product. The sign of the middle term determines whether the factorization is \((a + b)^2\) or \((a - b)^2\). Mastering this technique simplifies polynomial factoring and enhances problem-solving efficiency in algebra.

To factor the trinomial x² + 6xy + 9y² completely, we first recognize it as a perfect square trinomial, even though it contains two variables. A perfect square trinomial takes the form \(a^2 + 2ab + b^2\), which factors into \((a + b)^2\).

Here, the first term, x², is a perfect square because it can be written as x × x. The last term, 9y², is also a perfect square since it equals (3y) × (3y). The middle term, 6xy, matches the form 2ab, where a is x and b is 3y, because \$2 \times x \times 3y = 6xy$.

Since all conditions for a perfect square trinomial are met, the expression factors as:

This method highlights how recognizing perfect squares and verifying the middle term allows for efficient factoring of trinomials involving multiple variables.

To factor the trinomial 10a²b - 20ab² + 10b³ completely, start by identifying the greatest common factor (GCF) of all terms. Each term is divisible by 10 and contains at least one factor of b, so the GCF is 10b. Factoring out 10b simplifies the expression to a² - 2ab + b².

The remaining trinomial a² - 2ab + b² is a perfect square trinomial, recognizable by the pattern x² - 2xy + y² = (x - y)². Applying this formula, it factors into (a - b)². Therefore, the complete factorization of the original expression is 10b(a - b)².

This process highlights the importance of first extracting the greatest common factor to simplify expressions before applying special factoring formulas such as the perfect square trinomial. Recognizing these patterns allows for efficient and accurate factorization of polynomial expressions.

When multiplying polynomials, the Distributive Property is a fundamental technique that allows you to expand expressions by multiplying each term in one polynomial by every term in the other. For example, when multiplying (a + b) by the trinomial (a² - ab + b²), you distribute each term in the binomial across the trinomial. Multiplying a by each term results in a³, -a²b, and ab². Then, distributing b gives a²b, -ab², and b³. Combining like terms, the negative and positive a²b terms cancel out, as do the ab² terms, leaving the simplified result a³ + b³. This expression is known as the sum of cubes.

Similarly, when multiplying (a - b) by (a² + ab + b²), the process is the same but with a change in signs. Distributing a yields a³, a²b, and ab², while distributing -b results in -a²b, -ab², and -b³. Again, like terms cancel out, leaving the difference of cubes: a³ - b³.

These results highlight the important algebraic identities for the sum and difference of cubes, which are essential for factoring and simplifying polynomial expressions. Recognizing that a³ + b³ and a³ - b³ can be expanded into these specific polynomial products helps deepen understanding of polynomial multiplication and factoring techniques.

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 sums and differences of cubes lies in managing the signs within the factors. The acronym SOAP—Same, Opposite, Always Positive—helps remember the sign pattern: the sign in the binomial matches the original expression's sign (same), the sign before the middle term in the trinomial is opposite, and the last term in the trinomial is always positive.

For instance, consider factoring \( 27y^3 - 1 \). Recognize that \( 27y^3 = (3y)^3 \) and \( 1 = 1^3 \). Since this is a difference of cubes, apply the formula to get \( (3y - 1)( (3y)^2 + 3y \cdot 1 + 1^2 ) \), which simplifies to \( (3y - 1)(9y^2 + 3y + 1) \). This method ensures that any polynomial expressed as a sum or difference of cubes can be factored efficiently into a binomial multiplied by a trinomial, facilitating further algebraic manipulation and simplification.

To factor the polynomial two thirds x cubed minus two thirds y cubed, start by recognizing that both terms share a common factor. 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, identify the greatest common factor (GCF), which is two thirds, and factor it out first. This simplifies the expression to:

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

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, \(a = x\) and \(b = y\), so applying the formula gives:

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

This is the fully factored form of the original polynomial. Factoring out the GCF first simplifies the process and allows the difference of cubes formula to be applied effectively, demonstrating the importance of identifying common factors before applying special factoring formulas.

To factor the polynomial 16x³ + 54y³, start by identifying the greatest common factor (GCF). The coefficients 16 and 54 share a GCF of 2. Factoring out 2 gives:

\$16x^3 + 54y^3 = 2(8x^3 + 27y^3)\(

Now, observe that both 8 and 27 are perfect cubes since \)8 = 2^3\( and \)27 = 3^3\(. This means the expression inside the parentheses is 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:

\)8x^3 + 27y^3 = (2x + 3y)( (2x)^2 - (2x)(3y) + (3y)^2 )\(

Simplify the terms inside the second parentheses:

\)(2x)^2 = 4x^2\(

\)(2x)(3y) = 6xy\(

\)(3y)^2 = 9y^2\(

So the factorization becomes:

\)8x^3 + 27y^3 = (2x + 3y)(4x^2 - 6xy + 9y^2)\(

Including the GCF factored out earlier, the fully factored form of the original polynomial is:

\)16x^3 + 54y^3 = 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 like the sum of cubes. Recognizing perfect cubes and correctly applying the sum of cubes identity allows for efficient polynomial factorization.

To factor the polynomial 3y x³ − 81 y⁴, start by identifying the greatest common factor (GCF). Both terms contain the variable y and are divisible by 3, so the GCF is 3y. Factoring out 3y gives:

\$3y(x^3 - 27 y^3)\(

Inside the parentheses, the expression is a difference of cubes since both x³ and 27 y³ are perfect cubes. Recall the difference of cubes formula:

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

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

\)3y (x - 3y)(x^2 + 3xy + 9y^2)$

This fully factors the original polynomial by first extracting the GCF and then applying the difference of cubes factorization. Understanding how to identify and factor out the GCF simplifies complex polynomials and recognizing perfect cubes allows the use of the difference of cubes formula effectively.

To factor the polynomial (p + q)³ − 27r³, recognize that it represents a difference of cubes. The expression can be rewritten as (p + q)³ − (3r)³, where both p + q and 3r are perfect cubes. The difference of cubes formula is given by:

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

Applying this formula with a = p + q and b = 3r, the factorization becomes:

\[ (p + q - 3r) \left( (p + q)^2 + (p + q)(3r) + (3r)^2 \right) \]

Expanding the squared term and the products inside the parentheses, we get:

\[ (p + q - 3r) \left( p^2 + 2pq + q^2 + 3r(p + q) + 9r^2 \right) \]

Distributing the terms further:

\[ (p + q - 3r) \left( p^2 + 2pq + q^2 + 3pr + 3qr + 9r^2 \right) \]

This expression represents the fully factored form of the original polynomial. Understanding how to identify and apply the difference of cubes formula is essential for factoring cubic polynomials efficiently, especially when dealing with sums or differences of perfect cubes.