Factor the following using trial and error. 3 y 2 − 14 y + 8 3y^2-14y+8

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

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

( 3 y + 2 ) ( y − 4 ) \left(3y+2\right)\left(y-4\right)

( y − 4 ) ( 3 y + 2 ) \left(y-4\right)\left(3y+2\right)

Factor the following using trial and error. 8 a 2 − 22 a + 15 8a^2-22a+15

( 4 a − 5 ) ( 2 a − 3 ) \left(4a-5\right)\left(2a-3\right)

( 4 a − 5 ) ( 2 a + 3 ) \left(4a-5\right)\left(2a+3\right)

( 2 a + 3 ) ( 4 a − 5 ) \left(2a+3\right)\left(4a-5\right)

( 3 a − 2 ) ( 4 a − 5 ) \left(3a-2\right)\left(4a-5\right)

Factor completely. 15 a 2 + 25 a − 40 15a^2+25a-40

5 ( 3 a + 8 ) ( a − 1 ) 5\left(3a+8\right)\left(a-1\right)

( 3 a + 8 ) ( a − 1 ) \left(3a+8\right)\left(a-1\right)

5 ( 3 a − 8 ) ( a − 1 ) 5\left(3a-8\right)\left(a-1\right)

( 3 a + 8 ) ( a + 1 ) \left(3a+8\right)\left(a+1\right)

Factor completely. 2 x 2 + 9 x + 9 2x^2+9x+9

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

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

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

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

Factor completely. − 6 x 2 + 17 x + 3 -6x^2+17x+3

( 6 x + 1 ) ( − x + 3 ) \left(6x+1\right)\left(-x+3\right)

( 6 x + 1 ) ( x + 3 ) \left(6x+1\right)\left(x+3\right)

( 6 x − 1 ) ( − x + 3 ) \left(6x-1\right)\left(-x+3\right)

( 6 x − 1 ) ( x + 3 ) \left(6x-1\right)\left(x+3\right)

Factor the following polynomial 6 x 3 + 9 x 2 − 15 x 6x^3+9x^2-15x

3 x ( x − 1 ) ( 2 x + 5 ) 3x\left(x-1\right)\left(2x+5\right)

3 x ( x + 1 ) ( 2 x − 5 ) 3x\left(x+1\right)\left(2x-5\right)

3 x ( 2 x − 5 ) ( x − 1 ) 3x\left(2x-5\right)\left(x-1\right)

( x − 1 ) ( 2 x + 5 ) \left(x-1\right)\left(2x+5\right)

Factor the following polynomial 2 x 2 + 7 x y + 3 y 2 2x^2+7xy+3y^2

( 2 x + y ) ( x + 3 y ) \left(2x+y\right)\left(x+3y\right)

( x + 2 y ) ( 3 x + y ) \left(x+2y\right)\left(3x+y\right)

( 2 x − y ) ( x + 3 y ) \left(2x-y\right)\left(x+3y\right)

( x + 3 y ) ( 2 x + y ) \left(x+3y\right)\left(2x+y\right)

7. Factoring

Factoring Trinomials of the Form ax² + bx + c

7. Factoring

Factoring Trinomials of the Form ax² + bx + c: Videos & Practice Problems

Video Lessons

Topic summary

Factoring trinomials with a leading coefficient other than one involves methods like trial and error and the AC method. Trial and error uses the FOIL technique to test binomial factors where the first terms multiply to $a x^{2}$ and the last terms multiply to $c$ . The AC method rewrites the trinomial by splitting the middle term based on factors of $a × c$ that sum to $b$ , then applies factoring by grouping. These approaches efficiently factor polynomials by leveraging coefficients, terms, and exponents.

concept

Factoring Trinomials Using Trial and Error

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Factoring Trinomials Using Trial and Error Video Summary

Factoring trinomials with a leading coefficient other than one, such as 2x² or -8x², requires additional strategies beyond simple factoring. One effective approach is the trial and error method, which involves making educated guesses for binomial factors and testing them using the FOIL technique until the original trinomial is reconstructed.

To factor a trinomial like 2x² + 11x + 5, start by identifying pairs of factors for the first term and the last term. The first terms of the binomials must multiply to the leading term, 2x², so possible pairs include x and 2x. The last terms must multiply to the constant term, 5, so possible pairs are 1 and 5. By testing combinations such as (x + 1)(2x + 5) and (x + 5)(2x + 1) using FOIL, you find that (x + 5)(2x + 1) expands back to the original trinomial, confirming the correct factorization.

The FOIL method involves multiplying the First terms, Outside terms, Inside terms, and Last terms of the binomials and then combining like terms. This process ensures the factors multiply correctly to the original trinomial.

For more complex trinomials like 6x² + 19x - 7, begin by listing all factor pairs of the leading coefficient and the constant term, considering their signs carefully. For example, factors of 6x² include 3x and 2x, or x and 6x. Factors of -7 include 7 and -1 or -7 and 1. Then, create all possible binomial pairs from these factors.

Instead of testing every combination with FOIL, focus on the sum of the products of the outside and inside terms, which must equal the middle term, 19x. For example, testing (3x - 1)(2x + 7) involves calculating 3x × 7 = 21x and -1 × 2x = -2x, which sum to 19x, matching the middle term. This confirms the correct factorization without exhaustive FOIL expansion.

In summary, factoring trinomials with a leading coefficient other than one involves identifying factor pairs of the first and last terms, then using the FOIL method and the sum of the inner and outer products to find the correct binomial factors. This trial and error approach, combined with strategic testing, streamlines the factoring process for more complex polynomials.

Problem

Factor the following using trial and error.
$3 y^{2} - 14 y + 8$

$(y - 2) (3 y - 4)$

$(3 y + 2) (y - 4)$

$(3 y - 2) (y - 4)$

$(y - 4) (3 y + 2)$

Problem

Factor the following using trial and error.
$8 a^{2} - 22 a + 15$

$(4 a - 5) (2 a + 3)$

$(4 a - 5) (2 a - 3)$

$(2 a + 3) (4 a - 5)$

$(3 a - 2) (4 a - 5)$

example

Factoring Trinomials Using Trial and Error Example 1

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Factoring Trinomials Using Trial and Error Example 1 Video Summary

Factoring trinomials with two variables, such as x and y, follows a similar approach to factoring single-variable trinomials by using the trial and error method. The key is to identify the factors of the first term (the a term) and the last term (the c term) of the trinomial. For example, if the first term is 2x², its factors could be 1x and 2x. If the last term is 3y², its factors could be 1y and 3y.

Next, you create possible binomial pairs by combining these factors, such as (1x + 1y) and (2x + 3y). To verify if this pair is correct, focus on the middle terms generated by the FOIL method, specifically the sum of the products of the inside and outside terms. Multiply the outside terms: 1x × 3y = 3xy, and multiply the inside terms: 1y × 2x = 2xy. Adding these gives 3xy + 2xy = 5xy, which matches the middle term of the original trinomial.

Therefore, the trinomial can be factored as $(x + y)(2x + 3y)$. This method highlights the importance of understanding how the distributive property and the FOIL technique work together to factor trinomials involving multiple variables efficiently.

example

Factoring Trinomials Using Trial and Error Example 2

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Factoring Trinomials Using Trial and Error Example 2 Video Summary

To factor a trinomial completely, it is effective to first identify and factor out the greatest common factor (GCF). This simplifies the polynomial and makes subsequent factoring easier, especially when dealing with larger coefficients. For example, consider the trinomial 8x² - 28x + 16. Begin by finding the prime factorization of each term: 8x² = 2 × 2 × 2 × x × x, 28x = 2 × 2 × 7 × x, and 16 = 2 × 2 × 2 × 2. The common factors across all terms are two twos, so the GCF is 4. Factoring out 4 yields 4(2x² - 7x + 4).

Next, apply the trial and error method to factor the simplified trinomial 2x² - 7x + 4. Start by listing the factors of the leading coefficient (a = 2) and the constant term (c = 4). The factors of 2 are 2 and 1, while the factors of 4 include ±1, ±2, and ±4. The goal is to find binomial pairs whose product equals the trinomial. Focus on the middle term, which results from the sum of the products of the inside and outside terms in the FOIL method.

Testing pairs, consider (2x + 1)(x - 4). Multiplying the outside terms gives 2x × (-4) = -8x, and the inside terms give 1 × x = 1x. Adding these yields -7x, matching the middle term's coefficient. Therefore, the factorization of the trinomial is 4(2x + 1)(x - 4).

This process highlights the importance of factoring out the GCF first to simplify the polynomial, then using systematic trial and error with factor pairs to identify the correct binomial factors. Understanding prime factorization and the FOIL method is essential for efficient polynomial factoring.

Problem

Factor completely.
$15 a^{2} + 25 a - 40$

$(3 a + 8) (a - 1)$

$5 (3 a - 8) (a - 1)$

$(3 a + 8) (a + 1)$

$5 (3 a + 8) (a - 1)$

concept

Factoring Trinomials by Grouping (AC Method)

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Factoring Trinomials by Grouping (AC Method) Video Summary

Factoring trinomials can be simplified using the AC method, also known as the grouping method, which is a faster alternative to trial and error. This technique transforms a trinomial with three terms into a four-term polynomial, allowing the use of factoring by grouping. The process begins by identifying the greatest common factor (GCF) of the polynomial; if the GCF is 1, you proceed to the next step.

Next, multiply the leading coefficient a by the constant term c in the trinomial of the form $ax^2 + bx + c$. Then, find two numbers that multiply to $a \times c$ and add up to the middle coefficient b. For example, in the trinomial \$2x^2 + 11x + 5$, multiply $a = 2$ and $c = 5\( to get 10. The pair of numbers that multiply to 10 and add to 11 are 1 and 10.

Rewrite the middle term \)11x$ as the sum of these two terms: $x + 10x$. This changes the trinomial into four terms: $2x^2 + x + 10x + 5$. Now, apply factoring by grouping by pairing the first two terms and the last two terms separately. Factor out the GCF from each pair: from $2x^2 + x$, factor out $x$ to get $x(2x + 1)$; from $10x + 5$, factor out 5 to get $5(2x + 1)$. Since both groups contain the common binomial factor $(2x + 1)$, factor this out to obtain the final factored form: $(2x + 1)(x + 5)\(.

The AC method is effective for factoring any trinomial where two numbers can be found that multiply to \)a \times c$ and add to $b$. This approach leverages multiplication and addition relationships within the polynomial to simplify factoring, making it a valuable tool for solving quadratic expressions efficiently.

Problem

Factor completely.
$2 x^{2} + 9 x + 9$

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

$(2 x + 3) (x - 3)$

$(x + 3) (2 x - 3)$

$(x - 3) (2 x - 3)$

Problem

Factor completely.
$- 6 x^{2} + 17 x + 3$

$(6 x + 1) (x + 3)$

$(6 x - 1) (- x + 3)$

$(6 x + 1) (- x + 3)$

$(6 x - 1) (x + 3)$

example

Factoring Trinomials by Grouping (AC Method) Example 3

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Factoring Trinomials by Grouping (AC Method) Example 3 Video Summary

To factor a trinomial completely, start by identifying the greatest common factor (GCF) of all terms. In this case, each coefficient is even, and every term contains the variable y, so the GCF is 2y. Factoring out 2y simplifies the trinomial to x² + 5xy + 6y².

Next, focus on factoring the simplified trinomial. Since the coefficient of x² is 1, look for two numbers that multiply to the constant term (6y²) and add up to the middle coefficient (5y). The pairs to consider are (1y, 6y) and (2y, 3y). The pair (2y, 3y) works because 2y + 3y = 5y.

Thus, the trinomial factors as (x + 2y)(x + 3y). Combining this with the GCF factored out earlier, the complete factorization is 2y(x + 2y)(x + 3y).

This process highlights the importance of first extracting the greatest common factor to simplify complex expressions before applying factoring techniques such as finding two numbers that multiply to the constant term and add to the middle coefficient. This method is essential for efficiently factoring trinomials involving multiple variables and coefficients.

Problem

Factor the following polynomial
$6 x^{3} + 9 x^{2} - 15 x$

$3 x (x + 1) (2 x - 5)$

$3 x (x - 1) (2 x + 5)$

$3 x (2 x - 5) (x - 1)$

$(x - 1) (2 x + 5)$

Problem

Factor the following polynomial
$2 x^{2} + 7 x y + 3 y^{2}$

$(x + 2 y) (3 x + y)$

$(2 x + y) (x + 3 y)$

$(2 x - y) (x + 3 y)$

$(x + 3 y) (2 x + y)$

Here’s what students ask on this topic:

When the leading coefficient (a) in a trinomial $ax x^{2} + b x + c$ is not 1, you can use methods like trial and error or the AC method. Trial and error involves guessing binomial factors where the first terms multiply to $a x^{2}$ and the last terms multiply to $c$ . You test these guesses using the FOIL technique until you find the correct factors. The AC method involves multiplying $a$ and $c$ , then finding two numbers that multiply to $a × c$ and add to $b$ . You rewrite the middle term using these numbers, creating four terms, and then factor by grouping. Both methods help simplify factoring trinomials with leading coefficients other than 1.

The trial and error method for factoring trinomials involves making educated guesses for binomial factors. For a trinomial $ax x^{2} + b x + c$ , you find pairs of terms where the first terms multiply to $a x^{2}$ and the last terms multiply to $c$ . You then use the FOIL technique to multiply these binomials and check if the result matches the original trinomial. If it does not, you try different combinations until you find the correct factors. This method relies on pattern recognition and testing possible factor pairs systematically.

The AC method simplifies factoring by first multiplying the leading coefficient $a$ and the constant term $c$ in the trinomial $ax x^{2} + b x + c$ . Then, you find two numbers that multiply to $a × c$ and add to the middle coefficient $b$ . You rewrite the middle term as the sum of these two numbers times $x$ , creating four terms. Next, you factor by grouping, pairing terms to factor out common factors, and finally factor out the common binomial. This method reduces trial and error by using multiplication and addition properties to guide the factoring process.

To factor a trinomial $ax x^{2} + b x + c$ using the AC method, follow these steps: (1) Multiply $a$ and $c$ . (2) Find two numbers that multiply to $a × c$ and add to $b$ . (3) Rewrite the middle term $b x$ as the sum of these two numbers times $x$ , creating four terms. (4) Group the four terms into two pairs and factor out the greatest common factor from each pair. (5) Factor out the common binomial factor from the resulting expression. This process transforms the trinomial into a product of two binomials.

The FOIL technique helps verify if your binomial factors multiply back to the original trinomial. FOIL stands for First, Outside, Inside, Last, referring to the terms you multiply: (1) Multiply the first terms of each binomial. (2) Multiply the outside terms. (3) Multiply the inside terms. (4) Multiply the last terms. Then, add all these products together. If the sum matches the original trinomial $ax x^{2} + b x + c$ , your factors are correct. If not, you need to try different binomial pairs. This method ensures accuracy in factoring.

Your Beginning Algebra tutors

Factoring Trinomials of the Form ax² + bx + c: Videos & Practice Problems

Factoring Trinomials Using Trial and Error

Factoring Trinomials Using Trial and Error Video Summary

Factor the following using trial and error.
$3 y^{2} - 14 y + 8$

Factor the following using trial and error.
$8 a^{2} - 22 a + 15$

Factoring Trinomials Using Trial and Error Example 1

Factoring Trinomials Using Trial and Error Example 1 Video Summary

Factoring Trinomials Using Trial and Error Example 2

Factoring Trinomials Using Trial and Error Example 2 Video Summary

Factor completely.
$15 a^{2} + 25 a - 40$

Factoring Trinomials by Grouping (AC Method)

Factoring Trinomials by Grouping (AC Method) Video Summary

Factor completely.
$2 x^{2} + 9 x + 9$

Factor completely.
$- 6 x^{2} + 17 x + 3$

Factoring Trinomials by Grouping (AC Method) Example 3

Factoring Trinomials by Grouping (AC Method) Example 3 Video Summary

Factor the following polynomial
$6 x^{3} + 9 x^{2} - 15 x$

Factor the following polynomial
$2 x^{2} + 7 x y + 3 y^{2}$

Here’s what students ask on this topic:

How do you factor trinomials when the leading coefficient is not 1?

What is the trial and error method for factoring trinomials?

How does the AC method simplify factoring trinomials with a leading coefficient other than one?

What are the steps to factor a trinomial using the AC method?

How do you use the FOIL technique to check your factors?

Your Beginning Algebra tutors

Factoring Trinomials of the Form ax² + bx + c: Videos & Practice Problems

Factoring Trinomials Using Trial and Error

Factoring Trinomials Using Trial and Error Video Summary

Factor the following using trial and error.3y2−14y+83y^2-14y+8

Factor the following using trial and error. 8a2−22a+158a^2-22a+15

Factoring Trinomials Using Trial and Error Example 1

Factoring Trinomials Using Trial and Error Example 1 Video Summary

Factoring Trinomials Using Trial and Error Example 2

Factoring Trinomials Using Trial and Error Example 2 Video Summary

Factor completely.15a2+25a−4015a^2+25a-40

Factoring Trinomials by Grouping (AC Method)

Factoring Trinomials by Grouping (AC Method) Video Summary

Factor completely.2x2+9x+92x^2+9x+9

Factor completely.−6x2+17x+3-6x^2+17x+3

Factoring Trinomials by Grouping (AC Method) Example 3

Factoring Trinomials by Grouping (AC Method) Example 3 Video Summary

Factor the following polynomial6x3+9x2−15x6x^3+9x^2-15x

Factor the following polynomial2x2+7xy+3y22x^2+7xy+3y^2

Here’s what students ask on this topic:

How do you factor trinomials when the leading coefficient is not 1?

How do you factor trinomials when the leading coefficient is not 1?

What is the trial and error method for factoring trinomials?

What is the trial and error method for factoring trinomials?

How does the AC method simplify factoring trinomials with a leading coefficient other than one?

How does the AC method simplify factoring trinomials with a leading coefficient other than one?

What are the steps to factor a trinomial using the AC method?

What are the steps to factor a trinomial using the AC method?

How do you use the FOIL technique to check your factors?

How do you use the FOIL technique to check your factors?

Your Beginning Algebra tutors

Factor the following using trial and error.
$3 y^{2} - 14 y + 8$

Factor the following using trial and error.
$8 a^{2} - 22 a + 15$

Factor completely.
$15 a^{2} + 25 a - 40$

Factor completely.
$2 x^{2} + 9 x + 9$

Factor completely.
$- 6 x^{2} + 17 x + 3$

Factor the following polynomial
$6 x^{3} + 9 x^{2} - 15 x$

Factor the following polynomial
$2 x^{2} + 7 x y + 3 y^{2}$