Solve 9p^2r+73pr+70r

$$9 p ^ { 2 } r + 73 p r + 70 r$$

Factor

$$r\left(p+7\right)\left(9p+10\right)$$

Solution Steps

Factor out r.

$$r\left(9p^{2}+73p+70\right)$$

Consider 9p^{2}+73p+70. Factor the expression by grouping. First, the expression needs to be rewritten as 9p^{2}+ap+bp+70. To find a and b, set up a system to be solved.

$$a+b=73 ab=9\times 70=630$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 630.

$$1,630 2,315 3,210 5,126 6,105 7,90 9,70 10,63 14,45 15,42 18,35 21,30$$

Calculate the sum for each pair.

$$1+630=631 2+315=317 3+210=213 5+126=131 6+105=111 7+90=97 9+70=79 10+63=73 14+45=59 15+42=57 18+35=53 21+30=51$$

The solution is the pair that gives sum 73.

$$a=10 b=63$$

Rewrite 9p^{2}+73p+70 as \left(9p^{2}+10p\right)+\left(63p+70\right).

$$\left(9p^{2}+10p\right)+\left(63p+70\right)$$

Factor out p in the first and 7 in the second group.

$$p\left(9p+10\right)+7\left(9p+10\right)$$

Factor out common term 9p+10 by using distributive property.

$$\left(9p+10\right)\left(p+7\right)$$

Rewrite the complete factored expression.

$$r\left(9p+10\right)\left(p+7\right)$$

Evaluate

$$r\left(p+7\right)\left(9p+10\right)$$