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

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

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

this problem, we've been asked to factor this trinomial using our trial and error method. So our first step is to list out the factors of our a term, as well as the factors of our c term. For our a term, we have just the factors three and one. But for our c term, we have the factors four and two, as well as eight and one. Now before I go on to creating our possible binomial factor pairs, I want to point out that our middle term here is negative. And I know that if I multiply these positive values times these positive values, I'm only going to get out positive values. I also know that if I made both of them negative, I would also still get out positive values. So I need to make either three and one positive and our factors for eight negative to get out something negative, or I could switch it and have the first group of factors be negative and the second group of factors be positive to get out something negative. So let's go ahead and try this first option here, and we're going to make three and one positive, but all of our factors for eight will be negative. Now we can go ahead and create some potential binomial factor pairs. Using three and one, we could pair those with negative four and negative two. Now remember, because we have so many factors here, we're gonna have a lot of potential combinations of binomial factor pairs. So to speed things up, we're going to focus on just the middle two steps of our foil technique because those are the steps that give us the middle term of our trinomial. So multiplying our outside terms, we have three y times negative two is negative six y, and multiplying our inside terms, we have negative four times one y is negative four y, and those combine to make negative 10 y. Now we did get out some negative y values, which is what we were looking for, but we didn't get 14. So let's go ahead and try swapping our four and two values. So now we're going to test out three y minus two times one y minus four and see what that gives us. Multiplying our outside terms, we have negative 12 y, and multiplying the inside we have negative two y, which does combine to make negative 14 y. So that tells me that this trinomial can be factored into three y minus two times one y minus four. We could always check our answer here by using our foil technique to completely multiply these two binomials together.

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) ( 3 y − 2 ) ( y − 4 )

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

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

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

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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Multiple Choice

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

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

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

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

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

Verified step by step guidance

Identify the quadratic expression to factor: \$3y^2 - 14y + 8$.

List the factors of the first coefficient (3) and the constant term (8). For 3, the factors are 1 and 3; for 8, the factors are 1, 2, 4, and 8.

Set up two binomials in the form $(ay + b)(cy + d)$ where $a$ and $c$ multiply to 3, and $b$ and $d$ multiply to 8. Possible pairs for $(a, c)$ are (3, 1) or (1, 3), and for $(b, d)$ are (1, 8), (2, 4), (4, 2), or (8, 1), considering positive and negative signs.

Use trial and error by expanding each possible binomial pair using the distributive property (FOIL method): $ (ay + b)(cy + d) = acy^2 + (ad + bc)y + bd $, and check if the middle term matches $-14y$.

Once the correct pair is found that produces the middle term $-14y$ and the constant term \$8$, write the factored form. In this case, it is $(3y - 2)(y - 4)$.

8:18m

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