Factor. 2y^2 + 5y + 3 ____

Identify a, b, c: Identify a, b, and c in the quadratic expression 2y^2 + 5y + 3. Compare 2y^2 + 5y + 3 with ax^2 + bx + c. a = 2 b = 5 c = 3Find numbers and add: Find two numbers that multiply to a*c (which is 2*3=6) and add up to b (which is 5). The two numbers that satisfy these conditions are 2 and 3 because: 2 * 3 = 6 2 + 3 = 5Rewrite middle term: Rewrite the middle term of the quadratic expression using the two numbers found in Step 2. The expression 2y^2 + 5y + 3 can be rewritten by splitting the middle term into 2y and 3y: 2y^2 + 5y + 3 = 2y^2 + 2y + 3y + 3Factor by grouping: Factor by grouping. Group the first two terms together and the last two terms together: (2y^2 + 2y) + (3y + 3) Factor out the common factor from each group: 2y(y + 1) + 3(y + 1)Factor common binomial: Factor out the common binomial factor. Since both groups contain the factor (y + 1), factor it out: 2y(y + 1) + 3(y + 1) = (2y + 3)(y + 1)

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Algebra 1

Factor quadratics with other leading coefficients

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Q. Factor.

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2y^2 + 5y + 3

Identify $a$ , $b$ , $c$ : Identify $a$ , $b$ , and $c$ in the quadratic expression $2y^2 + 5y + 3$ . Compare $2y^2 + 5y + 3$ with $ax^2 + bx + c$ . $a = 2$ $b$ $0$ $b$ $1$
Find numbers and add: Find two numbers that multiply to $a*c$ (which is $2*3=6$ ) and add up to $b$ (which is $5$ ). $\newline$ The two numbers that satisfy these conditions are $2$ and $3$ because: $\newline$ $2 * 3 = 6$ $\newline$ $2 + 3 = 5$
Rewrite middle term: Rewrite the middle term of the quadratic expression using the two numbers found in Step $2$ . $\newline$ The expression $2y^2 + 5y + 3$ can be rewritten by splitting the middle term into $2y$ and $3y$ : $\newline$ $2y^2 + 5y + 3 = 2y^2 + 2y + 3y + 3$
Factor by grouping: Factor by grouping. $\newline$ Group the first two terms together and the last two terms together: $\newline$ $(2y^2 + 2y) + (3y + 3)$ $\newline$ Factor out the common factor from each group: $\newline$ $2y(y + 1) + 3(y + 1)$
Factor common binomial: Factor out the common binomial factor. $\newline$ Since both groups contain the factor $(y + 1)$ , factor it out: $\newline$ $2y(y + 1) + 3(y + 1) = (2y + 3)(y + 1)$