Solve 21a^{7}b^3:(-7a^5b^2)+(-3ab)(-2a)+6a^2b-(-3a)

$$21 a ^ { 7 } b ^ { 3 } : ( - 7 a ^ { 5 } b ^ { 2 } ) + ( - 3 a b ) ( - 2 a ) + 6 a ^ { 2 } b - ( - 3 a )$$

Factor

$$3a\left(3ab+1\right)$$

Solution Steps

Multiply a and a to get a^{2}.

$$factor(\frac{21a^{7}b^{3}}{-7a^{5}b^{2}}-3a^{2}b\left(-2\right)+6a^{2}b-\left(-3a\right))$$

Cancel out 7b^{2}a^{5} in both numerator and denominator.

$$factor(\frac{3ba^{2}}{-1}-3a^{2}b\left(-2\right)+6a^{2}b-\left(-3a\right))$$

Anything divided by -1 gives its opposite.

$$factor(-3ba^{2}-3a^{2}b\left(-2\right)+6a^{2}b-\left(-3a\right))$$

Multiply -3 and -2 to get 6.

$$factor(-3ba^{2}+6a^{2}b+6a^{2}b-\left(-3a\right))$$

Combine -3ba^{2} and 6a^{2}b to get 3ba^{2}.

$$factor(3ba^{2}+6a^{2}b-\left(-3a\right))$$

Combine 3ba^{2} and 6a^{2}b to get 9ba^{2}.

$$factor(9ba^{2}-\left(-3a\right))$$

The opposite of -3a is 3a.

$$factor(9ba^{2}+3a)$$

Factor out 3.

$$3\left(3ba^{2}+a\right)$$

Consider 3ba^{2}+a. Factor out a.

$$a\left(3ba+1\right)$$

Rewrite the complete factored expression.

$$3a\left(3ba+1\right)$$

Evaluate

$$3a\left(3ab+1\right)$$