Factor the polynomial by grouping − x 2 − 5 x + 7 x + 35 -x^2-5x+7x+35 − x 2 − 5 x + 7 x + 35

( − x + 7 ) ( x + 5 ) \left(-x+7\right)\left(x+5\right) ( − x + 7 ) ( x + 5 )

( x + 7 ) ( x + 5 ) \left(x+7\right)\left(x+5\right) ( x + 7 ) ( x + 5 )

( x − 7 ) ( x + 5 ) \left(x-7\right)\left(x+5\right) ( x − 7 ) ( x + 5 )

( x + 7 ) ( − x + 5 ) \left(x+7\right)\left(-x+5\right) ( x + 7 ) ( − x + 5 )

Factor the polynomial by grouping. 6 x 3 − 2 x 2 + 3 x − 1 6x^3-2x^2+3x-1 6 x 3 − 2 x 2 + 3 x − 1

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

2 x 2 ( 3 x − 1 ) 2x^2\left(3x-1\right) 2 x 2 ( 3 x − 1 )

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

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

Factor the polynomial using special product formulas. 25 x 2 − 110 x + 121 25x^2-110x+121 25 x 2 − 110 x + 121

( 5 x − 11 ) 2 \left(5x-11\right)^2 ( 5 x − 11 ) 2

( 5 x − 10 ) 2 \left(5x-10\right)^2 ( 5 x − 10 ) 2

( 5 x + 11 ) ( 5 x − 11 ) \left(5x+11\right)\left(5x-11\right) ( 5 x + 11 ) ( 5 x − 11 )

( 5 x + 11 ) 2 \left(5x+11\right)^2 ( 5 x + 11 ) 2

Factor the polynomial using special product formulas. x 2 49 − 9 \frac{x^2}{49}-9 49 x 2 − 9

( x 7 + 3 ) ( x 7 − 3 ) \left(\frac{x}{7}+3\right)\left(\frac{x}{7}-3\right) ( 7 x + 3 ) ( 7 x − 3 )

( x 7 − 3 ) 2 \left(\frac{x}{7}-3\right)^2 ( 7 x − 3 ) 2

( 7 x − 3 ) 2 \left(7x-3\right)^2 ( 7 x − 3 ) 2

( 7 x + 3 ) ( 7 x − 3 ) \left(7x+3\right)\left(7x-3\right) ( 7 x + 3 ) ( 7 x − 3 )

Factor the polynomial. x 2 − 13 x + 40 x^2-13x+40 x 2 − 13 x + 40

( x − 5 ) ( x − 8 ) \left(x-5\right)\left(x-8\right) ( x − 5 ) ( x − 8 )

( x + 5 ) ( x + 8 ) \left(x+5\right)\left(x+8\right) ( x + 5 ) ( x + 8 )

( x − 4 ) ( x − 10 ) \left(x-4\right)\left(x-10\right) ( x − 4 ) ( x − 10 )

( x + 4 ) ( x + 10 ) \left(x+4\right)\left(x+10\right) ( x + 4 ) ( x + 10 )

Factor the polynomial. x 2 − 2 x − 15 x^2-2x-15 x 2 − 2 x − 15

( x + 3 ) ( x − 5 ) \left(x+3\right)\left(x-5\right) ( x + 3 ) ( x − 5 )

( x − 3 ) ( x − 5 ) \left(x-3\right)\left(x-5\right) ( x − 3 ) ( x − 5 )

( x − 3 ) ( x + 5 ) \left(x-3\right)\left(x+5\right) ( x − 3 ) ( x + 5 )

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

Factor the polynomial. 4 x 2 − 19 x + 12 4x^2-19x+12 4 x 2 − 19 x + 12

( 4 x − 3 ) ( x − 4 ) \left(4x-3\right)\left(x-4\right) ( 4 x − 3 ) ( x − 4 )

( 4 x − 6 ) ( x − 2 ) \left(4x-6\right)\left(x-2\right) ( 4 x − 6 ) ( x − 2 )

( 2 x − 3 ) ( 2 x − 4 ) \left(2x-3\right)\left(2x-4\right) ( 2 x − 3 ) ( 2 x − 4 )

( 2 x − 6 ) ( 2 x − 2 ) \left(2x-6\right)\left(2x-2\right) ( 2 x − 6 ) ( 2 x − 2 )

Factor the polynomial. 3 x 2 − 2 x − 5 3x^2-2x-5 3 x 2 − 2 x − 5

( x + 1 ) ( 3 x − 5 ) \left(x+1\right)\left(3x-5\right) ( x + 1 ) ( 3 x − 5 )

( x + 3 ) ( x − 5 ) \left(x+3\right)\left(x-5\right) ( x + 3 ) ( x − 5 )

( x + 1 ) ( x − 5 ) \left(x+1\right)\left(x-5\right) ( x + 1 ) ( x − 5 )

( 3 x + 1 ) ( x − 5 ) \left(3x+1\right)\left(x-5\right) ( 3 x + 1 ) ( x − 5 )

0. Fundamental Concepts of Algebra

Factoring Polynomials

0. Fundamental Concepts of Algebra

Factoring Polynomials: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Factoring polynomials involves breaking down complex expressions into simpler factors. Key methods include identifying the greatest common factor (GCF) and using grouping for four-term polynomials. Special product formulas, such as the difference of squares and perfect square trinomials, aid in recognizing patterns for efficient factoring. The AC method is essential when the leading coefficient is not one, requiring the identification of factors that multiply to ac and add to b. Mastering these techniques enhances problem-solving skills in algebra.

concept

Introduction to Factoring Polynomials

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Introduction to Factoring Polynomials Video Summary

In algebra, the process of factoring is essential for simplifying expressions and solving equations. Factoring involves breaking down a complex polynomial into simpler components, which is the reverse of multiplication. For instance, if you have a polynomial like $2x^2 + 6$, you can factor it by identifying the greatest common factor (GCF).

The GCF is the largest expression that evenly divides each term in the polynomial. To find the GCF, you can use a method called a factor tree, which helps break down each term into its prime factors. For example, the term $6$ can be factored into $2 \times 3$, while $2x^2$ can be factored into $2 \times x \times x$. By analyzing both terms, you can identify the common factors. In this case, the GCF is $2$, as it is the only factor that appears in both terms.

Once the GCF is determined, you can factor it out of the polynomial. For $2x^2 + 6$, factoring out the $2$ gives you $2(x^2 + 3)$. To verify your factoring, you can use the distributive property to ensure that multiplying back results in the original expression.

Another example involves the polynomial $7x^2 - 5x$. Here, the factor tree shows that $7x^2$ breaks down to $7 \times x \times x$ and $5x$ breaks down to $5 \times x$. The common factor is $x$, which can be factored out, resulting in $x(7x - 5)$.

In more complex cases, such as $8x^3 + 16x$, recognizing that $8$ is a common factor simplifies the process. The GCF here is $8x$, leading to the factored form $8x(x^2 + 2)$. This method of factoring not only simplifies expressions but also aids in solving polynomial equations.

Understanding how to identify and factor out the GCF is a foundational skill in algebra that will be useful in various mathematical contexts, including polynomial division and solving quadratic equations.

Problem

Factor out the Greatest Common Factor in the polynomial.
$4x^2y-100y$

$y\left(4x^2-100\right)$

$4\left(x^2y-25y\right)$

$4y$ ( $x^2-25$ )

$4y\left(x^2-100\right)$

Problem

Factor out the Greatest Common Factor in the polynomial.
$-3x^4+12x^3-18x^2$

$3x\left(-x^2+4x-6\right)$

$3x^2\left(-x^2+4x-6\right)$

$3\left(-x^3+4x^2-6x\right)$

$3x^2\left(-3x^4+12x^3-18x^2\right)$

concept

Factor by Grouping

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Factor by Grouping Video Summary

Factoring polynomials can be approached in various ways, one of which is by identifying a greatest common factor (GCF). For instance, in the polynomial $x^3 - 2x^2$, the GCF is $x^2$, which can be factored out, resulting in $x^2(x - 2)$. However, not all polynomials will have a common factor across all terms, particularly when dealing with four-term expressions. In such cases, a method called grouping is often employed.

Grouping is particularly useful when you cannot find a single GCF for all terms. Typically, if you encounter a polynomial with four terms, it is a strong indicator that grouping may be the appropriate method. Additionally, you may notice that the coefficients of the terms are often multiples of each other, which can aid in the grouping process.

To apply the grouping method, first ensure that the polynomial is in standard form. For example, consider the polynomial $x^3 - 2x^2 + 4x + 8$. The next step is to group the terms into pairs, usually the first two and the last two terms. This can be represented as $(x^3 - 2x^2) + (4x + 8)$.

Next, factor out the GCF from each group. In the first group, $x^3 - 2x^2$, the GCF is $x^2$, leading to $x^2(x - 2)$. In the second group, $4x + 8$, the GCF is $4$, resulting in $4(x + 2)$. After factoring, you will have $x^2(x - 2) + 4(x + 2)$.

At this point, you may notice that both groups share a common factor. In this case, the common factor is $x - 2$. By factoring out $x - 2$, the expression can be rewritten as $(x - 2)(x^2 + 4)$. This represents the complete factorization of the polynomial.

In summary, when faced with a polynomial that has four terms and lacks a common factor, consider using the grouping method. By grouping the terms, factoring out the GCF from each group, and identifying any common factors, you can effectively simplify the polynomial. This method is particularly useful for polynomials where the coefficients are multiples of each other, leading to a successful factorization.

Problem

Factor the polynomial by grouping
$-x^2-5x+7x+35$

$\left(-x+7\right)\left(x+5\right)$

$\left(x+7\right)\left(x+5\right)$

$\left(x-7\right)\left(x+5\right)$

$\left(x+7\right)\left(-x+5\right)$

Problem

Factor the polynomial by grouping.
$6x^3-2x^2+3x-1$

$2x^2\left(3x-1\right)$

$\left(2x^2+x\right)\left(3x-1\right)$

$\left(2x+1\right)\left(3x^2-1\right)$

$\left(2x^2+1\right)\left(3x-1\right)$

concept

Factor Using Special Product Formulas

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Factor Using Special Product Formulas Video Summary

In algebra, recognizing special patterns in polynomial expressions is crucial for efficient factoring. One common pattern is the difference of squares, which can be expressed as $a^2 - b^2 = (a + b)(a - b)$. For example, the expression $x^2 - 36$ can be identified as a difference of squares since both $x^2$ and $36$ are perfect squares. Here, $a$ is $x$ and $b$ is $6$ (since $6^2 = 36$). Thus, the factored form is $(x + 6)(x - 6)$.

Another important pattern is the difference of cubes, represented by the formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. For instance, in the expression $x^3 - 27$, both $x^3$ and $27$ are perfect cubes, where $a$ is $x$ and $b$ is $3$ (since $3^3 = 27$). Applying the difference of cubes formula, we find the factors to be $(x - 3)(x^2 + 3x + 9)$.

Additionally, perfect square trinomials follow the pattern $a^2 + 2ab + b^2 = (a + b)^2$. For example, the expression $x^2 + 12x + 36$ can be analyzed by identifying $a$ as $x$ and $b$ as $6$ (since $6^2 = 36$). The middle term $12x$ confirms the pattern since $2 \cdot 6 \cdot x = 12x$. Therefore, this trinomial factors to $(x + 6)^2$.

Understanding these special product formulas allows for quick and accurate factoring of polynomials, making it easier to solve equations and simplify expressions. Always remember to verify your factorizations by expanding the factors back to the original expression.

Problem

Factor the polynomial using special product formulas.
$25x^2-110x+121$

$\left(5x-10\right)^2$

$\left(5x+11\right)\left(5x-11\right)$

$\left(5x+11\right)^2$

$\left(5x-11\right)^2$

Problem

Factor the polynomial using special product formulas.
$\frac{x^2}{49}-9$

$\left(\frac{x}{7}-3\right)^2$

$\left(\frac{x}{7}+3\right)\left(\frac{x}{7}-3\right)$

$\left(7x-3\right)^2$

$\left(7x+3\right)\left(7x-3\right)$

concept

Factor Using the AC Method When a Is 1

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Factor Using the AC Method When a Is 1 Video Summary

When factoring polynomials, particularly quadratic expressions in the form $ ax^2 + bx + c $, a systematic approach can simplify the process. One effective method is known as the AC method, which is particularly useful when the leading coefficient $ a $ is equal to 1. This method helps identify two numbers that multiply to $ c $ and add to $ b $.

To begin, ensure the polynomial is in standard form. For example, in the expression $ x^2 + 7x + 10 $, we identify $ a = 1 $, $ b = 7 $, and $ c = 10 $. The first step in the AC method is to calculate $ a \times c $, which in this case is $ 1 \times 10 = 10 $. Next, we list the pairs of factors of 10, considering both positive and negative combinations:

1 and 10
-1 and -10
2 and 5
-2 and -5

After listing the factors, the next step is to determine which pair adds up to $ b $. In our example, we check:

1 + 10 = 11 (not a match)
-1 - 10 = -11 (not a match)
2 + 5 = 7 (this works)
-2 - 5 = -7 (not a match)

Here, the numbers 2 and 5 multiply to 10 and add to 7, fulfilling the requirements. Thus, we can express the original polynomial as the product of two binomials:

$ (x + 2)(x + 5) $

To verify, we can expand this product using the distributive property (often referred to as the FOIL method):

$ (x + 2)(x + 5) = x^2 + 5x + 2x + 10 = x^2 + 7x + 10 $

This confirms that our factorization is correct. The AC method is a reliable technique for factoring quadratics, especially when the leading coefficient is 1, allowing for a clear and systematic approach to finding the necessary factors.

Problem

Factor the polynomial.
$x^2-13x+40$

$\left(x+5\right)\left(x+8\right)$

$\left(x-5\right)\left(x-8\right)$

$\left(x-4\right)\left(x-10\right)$

$\left(x+4\right)\left(x+10\right)$

Problem

Factor the polynomial.
$x^2-2x-15$

$\left(x-3\right)\left(x-5\right)$

$\left(x-3\right)\left(x+5\right)$

$\left(x+3\right)\left(x-5\right)$

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

concept

Factor Using the AC Method When a Is Not 1

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Factor Using the AC Method When a Is Not 1 Video Summary

Factoring polynomials can be straightforward when the leading coefficient (a) is equal to 1, but it becomes more complex when a is not equal to 1. In such cases, the AC method is a useful approach. This method involves several steps that help in breaking down the polynomial into manageable parts.

To illustrate, consider the polynomial 2x² + 7x + 6. The first step is to identify the values of a, b, and c, where a = 2, b = 7, and c = 6. The next step is to calculate the product of a and c, which in this case is 2 × 6 = 12. This product is crucial as it helps in finding the factors that will aid in the factoring process.

Next, list the factor pairs of 12: (1, 12), (2, 6), and (3, 4). The goal is to find a pair of factors that add up to the b term, which is 7. The pair (3, 4) meets this requirement since 3 + 4 = 7.

However, unlike when a = 1, simply substituting these factors into binomials will not yield the original polynomial. Instead, we need to rewrite the polynomial as a four-term expression: 2x² + 3x + 4x + 6. This step allows us to utilize factoring by grouping.

Now, group the terms: (2x² + 3x) + (4x + 6). From the first group, factor out the greatest common factor, which is x, resulting in x(2x + 3). From the second group, factor out 2, yielding 2(2x + 3). Now, we have:

x(2x + 3) + 2(2x + 3)

Since both groups contain the common factor (2x + 3), we can factor this out, leading to:

(2x + 3)(x + 2)

Thus, the original polynomial 2x² + 7x + 6 can be factored as (2x + 3)(x + 2). This method of grouping and factoring is essential when dealing with polynomials where the leading coefficient is not equal to 1, ensuring accurate results every time.

Problem

Factor the polynomial.
$4x^2-19x+12$

$\left(4x-3\right)\left(x-4\right)$

$\left(4x-6\right)\left(x-2\right)$

$\left(2x-3\right)\left(2x-4\right)$

$\left(2x-6\right)\left(2x-2\right)$

Problem

Factor the polynomial.
$3x^2-2x-5$

$\left(x+3\right)\left(x-5\right)$

$\left(x+1\right)\left(x-5\right)$

$\left(3x+1\right)\left(x-5\right)$

$\left(x+1\right)\left(3x-5\right)$

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Here’s what students ask on this topic:

What is the greatest common factor (GCF) in polynomial factoring?

The greatest common factor (GCF) in polynomial factoring is the largest expression that evenly divides all terms in the polynomial. To find the GCF, you break down each term into its prime factors and identify the common factors. For example, in the polynomial ${2x}^{2} + 6$ , the GCF is 2 because it is the largest number that divides both terms. Factoring out the GCF simplifies the polynomial, making it easier to work with. The process involves writing the polynomial as a product of the GCF and the remaining terms.

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How do you factor polynomials by grouping?

Factoring by grouping is used when a polynomial has four terms. The process involves grouping the terms into pairs and factoring out the GCF from each pair. For example, consider the polynomial $x^{3} - 2 x^{2} + 4x - 8$ . First, group the terms: $(x^{3} - 2 x^{2}) + (4x - 8)$ . Next, factor out the GCF from each group: $x x^{2} (x - 2) + 4(x - 2)$ . Finally, factor out the common binomial factor: $(x - 2)(x^{2} + 4)$ .

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What are special product formulas in polynomial factoring?

Special product formulas in polynomial factoring include patterns like the difference of squares and perfect square trinomials. The difference of squares formula is $a^{2} - b^{2} = (a + b)(a - b)$ . For example, $x^{2} - 36$ factors to $(x + 6)(x - 6)$ . Perfect square trinomials follow the pattern $a^{2} + 2ab + b^{2} = (a + b) (a + b)$ . For example, $x^{2} + 12x + 36$ factors to $(x + 6) (x + 6)$ .

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How does the AC method work for factoring polynomials?

The AC method is used for factoring polynomials of the form $ax 2 + bx + c$ when the leading coefficient $a$ is not 1. First, multiply $a$ and $c$ . List the factors of $ac$ and find the pair that adds to $b$ . Rewrite the middle term using these factors, then group and factor by grouping. For example, for $2x 2 + 7x + 6$ , multiply $2$ and $6$ to get $12$ . The factors $3$ and $4$ add to $7$ . Rewrite as $2x 2 + 3x + 4x + 6$ , then group: $x(2x + 3) + 2(2x + 3)$ . Factor out the common binomial: $(2x + 3)(x + 2)$ .

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What is the difference between factoring by grouping and the AC method?

Factoring by grouping is used for polynomials with four terms, where you group terms into pairs and factor out the GCF from each pair. For example, $x^{3} - 2 x^{2} + 4x - 8$ is grouped as $(x^{3} - 2 x^{2}) + (4x - 8)$ , then factored to $x (x - 2) + 4(x - 2)$ , and finally $(x - 2)(x 2 + 4)$ . The AC method is used for trinomials of the form $ax 2 + bx + c$ when $a$ is not 1. It involves multiplying $a$ and $c$ , finding factors that add to $b$ , rewriting the polynomial, and then grouping. For example, $2x 2 + 7x + 6$ is rewritten as $2x 2 + 3x + 4x + 6$ , grouped as $x(2x + 3) + 2(2x + 3)$ , and factored to $(2x + 3)(x + 2)$ .