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

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

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

Here, we've been asked to factor this trinomial using our trial and error method. Our first step is to list the factors of our a value as well as for our c value. So the factors of eight are four and two, as well as eight and one. And for 15, we have fifteen and one, as well as five and three. Now before I go on and list our possible binomial factor pairs, I want to point out that our middle term here is negative. And I know that multiplying these positive factors by these other positive factors is only going to produce a positive middle term. So in order to get something that's negative, I need either our first group of factors to be positive and our second group of factors to be negative, or vice versa, where this one is negative and this one is positive. So let's go ahead and try the first option and make our factors for 15 all negative. Now I can write out some possible binomial factor pairs. Starting with 4a and 2a, we could pair those with negative 15 and negative one. Now remember, when we're dealing with this many possible factors, we're going to have a lot of combinations of possible binomial factor pairs. So rather than foiling out every single one of these binomial factor pairs, we're going to focus on just the middle two steps because it's those two steps that create our middle term of the trinomial. So multiplying our outside terms, we have four a times negative one, which is negative 4a, and multiplying our inside terms, we have 15 times 2a, which is negative 30a. Summing those together, we get negative 34a, which is negative like we're looking for but much too large of a value. So next I could try swapping my fifteen and one factors, but doing a little quick mental math I can quickly see that that will lead to multiplying four a times 15, which is even larger than 34 a. So rather than doing this, I'm gonna go ahead and skip to my second factor pairs for 15, and we're gonna test out four a minus five times two a minus three. Multiplying the outside, we have 4a times negative three, which is negative 12a. And multiplying the inside, we have negative five times 2a, which is negative 10a. And those sum to negative 22 a, which is the term that we are looking for. So that means that we can factor this trinomial with the binomial factors of four a minus five times two a minus three.

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

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

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

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

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

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7. Factoring

Factoring Trinomials of the Form ax² + bx + c

Intermediate Algebra

7. Factoring

Factoring Trinomials of the Form ax² + bx + c

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객관식

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

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

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

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

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

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검증된 단계별 안내

Identify the quadratic expression to factor: $8a^2 - 22a + 15$.

Look for two binomials of the form $(pa + q)(ra + s)$ such that when multiplied, they give the original quadratic. Here, $p$ and $r$ multiply to $8$ (the coefficient of $a^2$), and $q$ and $s$ multiply to $15$ (the constant term).

List the factor pairs of 8: $(1, 8)$, $(2, 4)$, and the factor pairs of 15: $(1, 15)$, $(3, 5)$. Use these to form possible binomials.

Use trial and error by multiplying the binomials and checking if the middle term (the coefficient of $a$) matches $-22a$. Remember to consider the signs of $q$ and $s$ to get the correct middle term.

Once the correct pair is found, write the factored form as $(4a - 5)(2a - 3)$, confirming that the product expands back to the original quadratic.