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

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

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

For this problem, we've been asked to factor this trinomial completely, and we're going to use the method that we've been learning recently, called the AC method, or also known as the grouping method. So our first step is to identify our a value, our c value, as well as our b value. And next, we want to multiply our a and c value and find all of its factor pairs. So our a value is two, and our c value is nine. So multiplying those together, we get 18. And we wanna identify a pair of factors that multiply to 18 and also add to our b value of nine. So what are some factors of 18? Well, we have two and nine, which combine to make 11. We also have three and six, which combine to make nine. So that tells me that this is the factor pair that I am looking for. So now that we've identified our factor pair, we're going to split apart our middle term to turn this three term polynomial into a four term polynomial. So we can rewrite our polynomial as a two x squared, plus three x, plus six x, plus nine. And now we're going to apply factoring by grouping. So let's start by grouping our first two terms and last two terms and seeing if that works out. For our first two terms, our GCF is x, and for our last two terms, our GCF is three. So since both set of terms have a GCF, we can now continue our factoring by grouping by factoring them out. Factoring out x, we're left with two x plus three, and factoring out three, we are also left with two x plus three, which is perfect because now we have a common binomial that we can also factor out. So factoring out two x plus three, we're left with x plus three, and this here is our final answer.

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 )

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

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

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

7. Factoring

Factoring Trinomials of the Form ax² + bx + c

Beginning & Intermediate Algebra

7. Factoring

Factoring Trinomials of the Form ax² + bx + c

Struggling with Beginning & Intermediate Algebra?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

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

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

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

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

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

Verified step by step guidance

Identify the quadratic expression to factor: \$2x^{2} + 9x + 9$.

Multiply the coefficient of $x^{2}$ (which is 2) by the constant term (which is 9) to get \$2 \times 9 = 18$.

Find two numbers that multiply to 18 and add up to the middle coefficient, 9. These numbers are 6 and 3 because \$6 \times 3 = 18$ and $6 + 3 = 9$.

Rewrite the middle term \$9x$ as $6x + 3x$, so the expression becomes $2x^{2} + 6x + 3x + 9$.

Group the terms in pairs and factor each group: from \$2x^{2} + 6x$, factor out $2x$ to get $2x(x + 3)$; from $3x + 9$, factor out 3 to get $3(x + 3)$. Then factor out the common binomial $(x + 3)$ to get $(2x + 3)(x + 3)$.

8:18m

Watch next

Master Factoring Trinomials Using Trial and Error with a bite sized video explanation from Patrick Ford

Start learning