Solve 7r^2x-39rx-70x

$$7 r ^ { 2 } x - 39 r x - 70 x$$

Factor

$$x\left(r-7\right)\left(7r+10\right)$$

Solution Steps

Factor out x.

$$x\left(7r^{2}-39r-70\right)$$

Consider 7r^{2}-39r-70. Factor the expression by grouping. First, the expression needs to be rewritten as 7r^{2}+ar+br-70. To find a and b, set up a system to be solved.

$$a+b=-39 ab=7\left(-70\right)=-490$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -490.

$$1,-490 2,-245 5,-98 7,-70 10,-49 14,-35$$

Calculate the sum for each pair.

$$1-490=-489 2-245=-243 5-98=-93 7-70=-63 10-49=-39 14-35=-21$$

The solution is the pair that gives sum -39.

$$a=-49 b=10$$

Rewrite 7r^{2}-39r-70 as \left(7r^{2}-49r\right)+\left(10r-70\right).

$$\left(7r^{2}-49r\right)+\left(10r-70\right)$$

Factor out 7r in the first and 10 in the second group.

$$7r\left(r-7\right)+10\left(r-7\right)$$

Factor out common term r-7 by using distributive property.

$$\left(r-7\right)\left(7r+10\right)$$

Rewrite the complete factored expression.

$$x\left(r-7\right)\left(7r+10\right)$$

Evaluate

$$x\left(r-7\right)\left(7r+10\right)$$