Factor. 4d^2 + 7d + 3 ____

Identify coefficients: Identify the coefficients a, b, and c in the quadratic expression 4d^2 + 7d + 3 by comparing it to the standard form ax^2 + bx + c. a = 4, b = 7, c = 3.Find two numbers: Find two numbers that multiply to a*c (which is 4*3 = 12) and add up to b (which is 7). The two numbers that satisfy these conditions are 3 and 4 because 3*4 = 12 and 3+4 = 7.Rewrite middle term: Rewrite the middle term, 7d, using the two numbers found in the previous step. This will allow us to split the middle term into two terms. 4d^2 + 7d + 3 can be rewritten as 4d^2 + 4d + 3d + 3.Factor by grouping: Factor by grouping. Group the first two terms together and the last two terms together, then factor out the common factor from each group. From 4d^2 + 4d, we can factor out 4d, resulting in 4d(d + 1). From 3d + 3, we can factor out 3, resulting in 3(d + 1). Now we have 4d(d + 1) + 3(d + 1).Factor out common factor: Factor out the common binomial factor (d + 1) from both groups. The expression becomes (4d + 3)(d + 1).

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Factor. $\newline$ $4d^2 + 7d + 3$

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Algebra 1

Factor quadratics with other leading coefficients

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Q. Factor.

\newline

4d^2 + 7d + 3

Identify coefficients: Identify the coefficients $a$ , $b$ , and $c$ in the quadratic expression $4d^2 + 7d + 3$ by comparing it to the standard form $ax^2 + bx + c$ .
$a = 4$ , $b = 7$ , $c = 3$ .
Find two numbers: Find two numbers that multiply to $a*c$ (which is $4*3 = 12$ ) and add up to $b$ (which is $7$ ). $\newline$ The two numbers that satisfy these conditions are $3$ and $4$ because $3*4 = 12$ and $3+4 = 7$ .
Rewrite middle term: Rewrite the middle term, $7d$ , using the two numbers found in the previous step. This will allow us to split the middle term into two terms. $\newline$ $4d^2 + 7d + 3$ can be rewritten as $4d^2 + 4d + 3d + 3$ .
Factor by grouping: Factor by grouping. Group the first two terms together and the last two terms together, then factor out the common factor from each group. $\newline$ From $4d^2 + 4d$ , we can factor out $4d$ , resulting in $4d(d + 1)$ . $\newline$ From $3d + 3$ , we can factor out $3$ , resulting in $3(d + 1)$ . $\newline$ Now we have $4d(d + 1) + 3(d + 1)$ .
Factor out common factor: Factor out the common binomial factor $(d + 1)$ from both groups. $\newline$ The expression becomes $(4d + 3)(d + 1)$ .