Solve (35x^2+32x-99)

$$( 35 x ^ { 2 } + 32 x - 99 )$$

Factor

$$\left(7x-9\right)\left(5x+11\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 35x^{2}+ax+bx-99. To find a and b, set up a system to be solved.

$$a+b=32 ab=35\left(-99\right)=-3465$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -3465.

$$-1,3465 -3,1155 -5,693 -7,495 -9,385 -11,315 -15,231 -21,165 -33,105 -35,99 -45,77 -55,63$$

Calculate the sum for each pair.

$$-1+3465=3464 -3+1155=1152 -5+693=688 -7+495=488 -9+385=376 -11+315=304 -15+231=216 -21+165=144 -33+105=72 -35+99=64 -45+77=32 -55+63=8$$

The solution is the pair that gives sum 32.

$$a=-45 b=77$$

Rewrite 35x^{2}+32x-99 as \left(35x^{2}-45x\right)+\left(77x-99\right).

$$\left(35x^{2}-45x\right)+\left(77x-99\right)$$

Factor out 5x in the first and 11 in the second group.

$$5x\left(7x-9\right)+11\left(7x-9\right)$$

Factor out common term 7x-9 by using distributive property.

$$\left(7x-9\right)\left(5x+11\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$35x^{2}+32x-99=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-32±\sqrt{32^{2}-4\times 35\left(-99\right)}}{2\times 35}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 35\left(-99\right)}}{2\times 35}$$

Multiply -4 times 35.

$$x=\frac{-32±\sqrt{1024-140\left(-99\right)}}{2\times 35}$$

Multiply -140 times -99.

$$x=\frac{-32±\sqrt{1024+13860}}{2\times 35}$$

Add 1024 to 13860.

$$x=\frac{-32±\sqrt{14884}}{2\times 35}$$

Take the square root of 14884.

$$x=\frac{-32±122}{2\times 35}$$

Multiply 2 times 35.

$$x=\frac{-32±122}{70}$$

Now solve the equation x=\frac{-32±122}{70} when ± is plus. Add -32 to 122.

$$x=\frac{90}{70}$$

Reduce the fraction \frac{90}{70} to lowest terms by extracting and canceling out 10.

$$x=\frac{9}{7}$$

Now solve the equation x=\frac{-32±122}{70} when ± is minus. Subtract 122 from -32.

$$x=-\frac{154}{70}$$

Reduce the fraction \frac{-154}{70} to lowest terms by extracting and canceling out 14.

$$x=-\frac{11}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{7} for x_{1} and -\frac{11}{5} for x_{2}.

$$35x^{2}+32x-99=35\left(x-\frac{9}{7}\right)\left(x-\left(-\frac{11}{5}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$35x^{2}+32x-99=35\left(x-\frac{9}{7}\right)\left(x+\frac{11}{5}\right)$$

Subtract \frac{9}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$35x^{2}+32x-99=35\times \frac{7x-9}{7}\left(x+\frac{11}{5}\right)$$

Add \frac{11}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$35x^{2}+32x-99=35\times \frac{7x-9}{7}\times \frac{5x+11}{5}$$

Multiply \frac{7x-9}{7} times \frac{5x+11}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$35x^{2}+32x-99=35\times \frac{\left(7x-9\right)\left(5x+11\right)}{7\times 5}$$

Multiply 7 times 5.

$$35x^{2}+32x-99=35\times \frac{\left(7x-9\right)\left(5x+11\right)}{35}$$

Cancel out 35, the greatest common factor in 35 and 35.

$$35x^{2}+32x-99=\left(7x-9\right)\left(5x+11\right)$$

Evaluate

$$\left(7x-9\right)\left(5x+11\right)$$