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

Identify a, b, c: Identify a, b, and c in the quadratic expression 4u^2 + 7u + 3. Compare 4u^2 + 7u + 3 with the standard form ax^2 + bx + c. a = 4 b = 7 c = 3Find Factors and Sum: Find two numbers that multiply to a*c (which is 4*3=12) and add up to b (which is 7). We need to find two numbers that satisfy these conditions. After checking possible pairs of factors of 12, we find that 3 and 4 are the numbers we are looking for because 3*4 = 12 and 3 + 4 = 7.Rewrite Middle Term: Rewrite the middle term 7u using the two numbers 3 and 4 found in Step 2. 4u^2 + 7u + 3 can be rewritten as 4u^2 + 4u + 3u + 3.Factor by Grouping: Factor by grouping. Group the first two terms and the last two terms separately. (4u^2 + 4u) + (3u + 3) Factor out the common factor from each group. 4u(u + 1) + 3(u + 1)Factor out Common Binomial: Factor out the common binomial factor (u + 1). The expression now becomes (4u + 3)(u + 1).

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Algebra 1

Factor quadratics with other leading coefficients

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Q. Factor.

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4u^2 + 7u + 3

Identify $a$ , $b$ , $c$ : Identify $a$ , $b$ , and $c$ in the quadratic expression $4u^2 + 7u + 3$ . Compare $4u^2 + 7u + 3$ with the standard form $ax^2 + bx + c$ . $a = 4$ $b$ $0$ $b$ $1$
Find Factors and Sum: Find two numbers that multiply to $a*c$ (which is $4*3=12$ ) and add up to $b$ (which is $7$ ). $\newline$ We need to find two numbers that satisfy these conditions. $\newline$ After checking possible pairs of factors of $12$ , we find that $3$ and $4$ are the numbers we are looking for because $3*4 = 12$ and $3 + 4 = 7$ .
Rewrite Middle Term: Rewrite the middle term $7u$ using the two numbers $3$ and $4$ found in Step $2$ . $\newline$ $4u^2 + 7u + 3$ can be rewritten as $4u^2 + 4u + 3u + 3$ .
Factor by Grouping: Factor by grouping. $\newline$ Group the first two terms and the last two terms separately. $\newline$ $(4u^2 + 4u) + (3u + 3)$ $\newline$ Factor out the common factor from each group. $\newline$ $4u(u + 1) + 3(u + 1)$
Factor out Common Binomial: Factor out the common binomial factor $(u + 1)$ . The expression now becomes $(4u + 3)(u + 1)$ .