Factor. 4w^2 + 8w + 3 ______

Identify a, b, c: Identify a, b, and c in the quadratic expression 4w^2 + 8w + 3 by comparing it with the standard form ax^2 + bx + c. a = 4 b = 8 c = 3Find two numbers: Find two numbers that multiply to a*c (which is 4*3=12) and add up to b (which is 8). After trying different combinations, we find that the numbers 2 and 6 work because: 2 * 6 = 12 2 + 6 = 8Rewrite middle term: Rewrite the middle term of the quadratic expression using the two numbers found in the previous step. 4w^2 + 8w + 3 can be rewritten as 4w^2 + 2w + 6w + 3.Group terms and factor: Group the terms in pairs and factor out the common factors from each pair. From the first pair (4w^2 + 2w), factor out 2w: 2w(2w + 1) From the second pair (6w + 3), factor out 3: 3(2w + 1)Factor out common binomial: Notice that both groups have a common binomial factor (2w + 1). Factor out the common binomial factor to write the expression in factored form. The factored form is (2w + 1)(2w + 3).

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Algebra 1

Factor quadratics with other leading coefficients

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Q. Factor.

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4w^2 + 8w + 3

Identify $a$ , $b$ , $c$ : Identify $a$ , $b$ , and $c$ in the quadratic expression $4w^2 + 8w + 3$ by comparing it with the standard form $ax^2 + bx + c$ . $\newline$ $a = 4$ $\newline$ $b = 8$ $\newline$ $b$ $0$
Find two numbers: Find two numbers that multiply to $a*c$ (which is $4*3=12$ ) and add up to $b$ (which is $8$ ). $\newline$ After trying different combinations, we find that the numbers $2$ and $6$ work because: $\newline$ $2 \times 6 = 12$ $\newline$ $2 + 6 = 8$
Rewrite middle term: Rewrite the middle term of the quadratic expression using the two numbers found in the previous step. $\newline$ $4w^2 + 8w + 3$ can be rewritten as $4w^2 + 2w + 6w + 3$ .
Group terms and factor: Group the terms in pairs and factor out the common factors from each pair. $\newline$ From the first pair $4w^2 + 2w$ , factor out $2w$ : $\newline$ $2w(2w + 1)$ $\newline$ From the second pair $6w + 3$ , factor out $3$ : $\newline$ $3(2w + 1)$
Factor out common binomial: Notice that both groups have a common binomial factor $2w + 1$ . Factor out the common binomial factor to write the expression in factored form. $\newline$ The factored form is $2w + 1$ $2w + 3$ .