Factor the following trinomials completely. y 2 − 7 y + 12 y^2-7y+12

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

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

( y + 7 ) ( y − 1 ) \left(y+7\right)\left(y-1\right) ( y + 7 ) ( y − 1 )

( y − 1 ) ( y − 12 ) \left(y-1\right)\left(y-12\right) ( y − 1 ) ( y − 12 )

Factor the following trinomials completely. z 2 − 11 z + 30 z^2-11z+30

( z − 5 ) ( z − 6 ) \left(z-5\right)\left(z-6\right) ( z − 5 ) ( z − 6 )

( z + 5 ) ( z − 6 ) \left(z+5\right)\left(z-6\right) ( z + 5 ) ( z − 6 )

( z − 3 ) ( z − 10 ) \left(z-3\right)\left(z-10\right) ( z − 3 ) ( z − 10 )

( z + 5 ) ( z + 6 ) \left(z+5\right)\left(z+6\right) ( z + 5 ) ( z + 6 )

7. Factoring

Factoring Trinomials of the Form x² + bx + c

7. Factoring

Factoring Trinomials of the Form x² + bx + c: Videos & Practice Problems

Video Lessons

Topic summary

Factoring a trinomial of the form $x^{2} + b x + c$ involves finding two binomials whose product equals the original polynomial. Using the FOIL method, the constant term $c$ is the product of two constants, and the coefficient $b$ is their sum. By identifying factor pairs of $c$ that add to $b$ , students can factor trinomials efficiently. This technique applies when the coefficient of $x^2$ is one, reinforcing understanding of polynomials, coefficients, and binomials in standard form.

concept

Factoring Trinomials of the Form x² + bx + c

Video duration:

Factoring Trinomials of the Form x² + bx + c Video Summary

Factoring trinomials of the form x² + bx + c is a fundamental skill in algebra that allows you to rewrite quadratic expressions as the product of two binomials. This process is essentially the reverse of multiplying binomials using the FOIL method, which stands for First, Outside, Inside, Last. When multiplying two binomials like (x + p)(x + q), the result is a trinomial where the coefficient of x² is 1, the middle term’s coefficient b is the sum of p and q, and the constant term c is the product of p and q. This relationship is key to factoring.

To factor a trinomial x² + bx + c, you need to find two numbers that multiply to c and add up to b. For example, factoring x² + 10x + 21 involves finding two numbers that multiply to 21 and add to 10, which are 3 and 7. Thus, the factorization is (x + 3)(x + 7).

When the numbers are less straightforward, organizing factor pairs in a T-chart can help. For instance, to factor x² + 3x − 28, identify factor pairs of −28 and check which pair sums to 3. The pairs (−4, 7) multiply to −28 and add to 3, so the factorization is (x − 4)(x + 7).

Similarly, for x² − 11x + 30, factor pairs of 30 are considered. Since the middle term is negative, both factors must be negative to produce a positive product. The pair (−5, −6) multiplies to 30 and adds to −11, so the factorization is (x − 5)(x − 6).

This method only applies when the coefficient of x² is 1. To verify your factorization, multiply the binomials back using FOIL to ensure you return to the original trinomial. Mastering this technique enhances your ability to solve quadratic equations and simplifies expressions efficiently.

In summary, factoring trinomials involves identifying two numbers that satisfy both the product and sum conditions related to the constant and linear coefficients. Using systematic approaches like T-charts can simplify the process, especially with more complex numbers, reinforcing a deeper understanding of polynomial structure and algebraic manipulation.

Problem

Factor the following trinomials completely.
$y^{2} - 7 y + 12$

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

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

$\left(y+7\right)\left(y-1\right)$

$\left(y-1\right)\left(y-12\right)$

Problem

Factor the following trinomials completely.
$z^{2} - 11 z + 30$

$\left(z+5\right)\left(z-6\right)$

$\left(z-3\right)\left(z-10\right)$

$\left(z+5\right)\left(z+6\right)$

$\left(z-5\right)\left(z-6\right)$

example

Factoring Trinomials of the Form x² + bx + c Example 1

Video duration:

Factoring Trinomials of the Form x² + bx + c Example 1 Video Summary

When factoring a trinomial with a leading term of x², the method involves finding two numbers that multiply to the constant term (the c value) and add to the coefficient of the middle term (the b value). This approach remains effective even when the trinomial contains multiple variables, such as x and y.

Consider a trinomial like x² - 5xy + 6y². To factor this, first identify pairs of factors of the constant term, which is 6y². Possible factor pairs include 1y and 6y, or 2y and 3y. Since the middle term is -5xy, the two numbers must add to -5y. The pair -2y and -3y satisfies this because (-2y) + (-3y) = -5y, and their product is 6y².

Thus, the trinomial factors as:

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

To verify this factorization, apply the FOIL method (First, Outer, Inner, Last):

First: $x \times x = x^2$
Outer: $x \times (-3y) = -3xy$
Inner: $(-2y) \times x = -2xy$
Last: $(-2y) \times (-3y) = 6y^2$

Combining like terms, the expression becomes:

\[x^2 - 3xy - 2xy + 6y^2 = x^2 - 5xy + 6y^2\]

This matches the original trinomial, confirming the factorization is correct. This process highlights how factoring techniques extend naturally to polynomials with multiple variables by treating variable terms as part of the factors and carefully considering their coefficients.

Here’s what students ask on this topic:

To factor a trinomial of the form $x^{2} + b x + c$ , you need to find two numbers that multiply to $c$ and add to $b$ . These two numbers will be the constants in the binomials. The trinomial can then be factored as $(x + p) (x + q)$ , where $p$ and $q$ are the numbers found. This method works only when the coefficient of $x^{2}$ is 1. For example, to factor $x^{2} + 10 x + 21$ , find two numbers that multiply to 21 and add to 10, which are 3 and 7. So, the factorization is $(x + 3) (x + 7)$ .

First, identify the values of $b$ and $c$ : here, $b = 3$ and $c = -28$ . Next, list all factor pairs of $-28$ : $1 imes -28, -1 imes 28, 2 imes -14, -2 imes 14, 4 imes -7, -4 imes 7$ . Then, find the pair whose sum equals $3$ . Checking sums, $-4 + 7 = 3$ fits. Therefore, the trinomial factors as $(x - 4) (x + 7)$ . This method uses the distributive property in reverse, ensuring the product of constants equals $c$ and their sum equals $b$ .

The factoring method for trinomials of the form $x^{2} + b x + c$ relies on the coefficient of $x^{2}$ being 1 because it simplifies the relationship between the constants in the binomials. When the coefficient is 1, the product of the constants in the binomials equals $c$ , and their sum equals $b$ . If the coefficient is not 1, the product and sum relationships become more complex, requiring different factoring techniques such as factoring by grouping or using the quadratic formula. Thus, this method is specifically designed for monic trinomials where the leading coefficient is 1.

To verify your factorization, multiply the binomials back together using the FOIL method: multiply the First terms, Outside terms, Inside terms, and Last terms. For example, if you factored $x^{2} + 10 x + 21$ as $(x + 3) (x + 7)$ , multiply to get $x^{2} + 7 x + 3 x + 21$ , which simplifies to $x^{2} + 10 x + 21$ . If the product matches the original trinomial, your factorization is correct.

A systematic approach involves creating a T-chart to list all factor pairs of $c$ , including both positive and negative pairs. For each pair, calculate their sum and compare it to $b$ . The pair whose sum equals $b$ is used for factoring. For example, for $x^{2} + 3 x - 28$ , list factor pairs of -28: $(1, -28), (-1, 28), (2, -14), (-2, 14), (4, -7), (-4, 7)$ . Check sums until you find $-4 + 7 = 3$ . This method helps organize your work and reduces errors, especially with larger numbers or negative values.