Factor (6a^2+11a-10)÷(3a-2)

$(6 a^{2} + 11 a - 10) \div (3 a - 2)$

Step 1

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 6 \cdot - 10 = - 60$ and whose sum is $b = 11$ .

Factor $11$ out of $11 a$ .

$(6 a^{2} + 11 (a) - 10) \div (3 a - 2)$

Rewrite $11$ as $- 4$ plus $15$

$(6 a^{2} + (- 4 + 15) a - 10) \div (3 a - 2)$

Apply the distributive property.

$(6 a^{2} - 4 a + 15 a - 10) \div (3 a - 2)$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$((6 a^{2} - 4 a) + 15 a - 10) \div (3 a - 2)$

Factor out the greatest common factor (GCF) from each group.

$(2 a (3 a - 2) + 5 (3 a - 2)) \div (3 a - 2)$

Factor the polynomial by factoring out the greatest common factor, $3 a - 2$ .

$(3 a - 2) (2 a + 5) \div (3 a - 2)$

Step 2

Reduce the expression by cancelling the common factors.

Reduce the expression $(3 a - 2) (2 a + 5) \div (3 a - 2)$ by cancelling the common factors.

Cancel the common factor.

$(3 a - 2) (2 a + 5) \div (3 a - 2)$

Rewrite the expression.

$\frac{2 a + 5}{1}$

Divide $2 a + 5$ by $1$ .

$2 a + 5$

Factor (6a^2+11a-10)÷(3a-2)

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