Solve (7div7^3)*p^2+pq

$$( 7 \div 7 ^ { 3 } ) \cdot p ^ { 2 } + p q$$

Evaluate

$$pq+\frac{p^{2}}{49}$$

Solution Steps

Rewrite 7^{3} as 7\times 7^{2}. Cancel out 7 in both numerator and denominator.

$$\frac{1}{7^{2}}p^{2}+pq$$

Calculate 7 to the power of 2 and get 49.

$$\frac{1}{49}p^{2}+pq$$

Factor

$$\frac{p\left(p+49q\right)}{49}$$

Solution Steps

Rewrite 7^{3} as 7\times 7^{2}. Cancel out 7 in both numerator and denominator.

$$factor(\frac{1}{7^{2}}p^{2}+pq)$$

Calculate 7 to the power of 2 and get 49.

$$factor(\frac{1}{49}p^{2}+pq)$$

Factor out \frac{1}{49}.

$$\frac{p^{2}+49pq}{49}$$

Consider p^{2}+49pq. Factor out p.

$$p\left(p+49q\right)$$

Rewrite the complete factored expression.

$$\frac{p\left(p+49q\right)}{49}$$