Factor. 14tu + 7t + 10u + 5 ______

Identify Factors: Identify common factors in pairs of terms. We can group the terms as follows: (14tu + 7t) + (10u + 5). Now, we look for common factors in each group. In the first group (14tu + 7t), the common factor is 7t. In the second group (10u + 5), the common factor is 5.Factor Out Common Factors: Factor out the common factors from each group. From the first group, we factor out 7t, giving us 7t(2u + 1). From the second group, we factor out 5, giving us 5(2u + 1). Now we have 7t(2u + 1) + 5(2u + 1).Factor Out Binomial Factor: Factor out the common binomial factor. We notice that (2u + 1) is a common factor in both terms. We can factor (2u + 1) out of the expression, giving us (2u + 1)(7t + 5).

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Algebra 1

Factor polynomials

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Q. Factor.

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14tu + 7t + 10u + 5

Identify Factors: Identify common factors in pairs of terms. $\newline$ We can group the terms as follows: $14tu + 7t$ + $10u + 5$ . $\newline$ Now, we look for common factors in each group. $\newline$ In the first group $14tu + 7t$ , the common factor is $7t$ . $\newline$ In the second group $10u + 5$ , the common factor is $5$ .
Factor Out Common Factors: Factor out the common factors from each group. $\newline$ From the first group, we factor out $7t$ , giving us $7t(2u + 1)$ . $\newline$ From the second group, we factor out $5$ , giving us $5(2u + 1)$ . $\newline$ Now we have $7t(2u + 1) + 5(2u + 1)$ .
Factor Out Binomial Factor: Factor out the common binomial factor. $\newline$ We notice that $(2u + 1)$ is a common factor in both terms. $\newline$ We can factor $(2u + 1)$ out of the expression, giving us $(2u + 1)(7t + 5)$ .