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

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

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

Here we've been asked to factor this trinomial completely. And to do this, we're going to use the method that we've been practicing recently called the AC method, also known as the grouping method. So our first step here is to identify our a and c value as well as our b value. And our goal here is to find two numbers that multiply to our a times c value and also add to our b value. So let's take a look at a times c. We have negative six times three, which equals negative 18. So we want two numbers that multiply to negative 18 and add to our b value of 17. So let's test out some numbers. Well, I know negative two and nine make negative 18, but those two add to positive seven. So that's not what we're looking for. I also know that negative one times 18 makes negative 18, and those two do add to positive 17. So this is the factor pair that I'm looking for. Next, we're going to take the middle term in our trinomial and break it apart into the sum of these factor pairs. So we can rewrite this three term polynomial as a four term polynomial. So doing that we have negative 6x squared minus 1x plus 18x plus three. And now that we have a four term polynomial we can apply factoring by grouping. So let's take a look at grouping our first two terms and our last two terms. Do these pairings make any greatest common factors? For our first two terms, we have a greatest common factor of negative x. And for our last two terms, we have a greatest common factor of three. So we can go ahead and factor out these greatest common factors from these pairs. Factoring out negative x, we're left with six x plus one. And factoring out three, we are also left with six x plus one, which is perfect because now we have a common binomial that we can also factor out. Factoring out six x plus one, we're left with negative x plus three. And this here is our final answer.

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 )

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

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

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

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

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

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

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

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

Verified step by step guidance

Start with the quadratic expression: $-6x^2 + 17x + 3$.

Factor out the negative sign to make the leading coefficient positive: $-(6x^2 - 17x - 3)$.

Identify two numbers that multiply to $6 \times (-3) = -18$ and add to $-17$; these numbers will help split the middle term.

Rewrite the middle term using these two numbers and then group terms to factor by grouping.

Factor out the common binomial factor from each group to express the quadratic as a product of two binomials, then reapply the negative sign if factored out initially.

8:18m

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