Solve 35x^2+38x-45

$$35 { x }^{ 2 } +38x-45$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 35x^{2}+ax+bx-45. To find a and b, set up a system to be solved.

$$a+b=38 ab=35\left(-45\right)=-1575$$

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 -1575.

$$-1,1575 -3,525 -5,315 -7,225 -9,175 -15,105 -21,75 -25,63 -35,45$$

Calculate the sum for each pair.

$$-1+1575=1574 -3+525=522 -5+315=310 -7+225=218 -9+175=166 -15+105=90 -21+75=54 -25+63=38 -35+45=10$$

The solution is the pair that gives sum 38.

$$a=-25 b=63$$

Rewrite 35x^{2}+38x-45 as \left(35x^{2}-25x\right)+\left(63x-45\right).

$$\left(35x^{2}-25x\right)+\left(63x-45\right)$$

Factor out 5x in the first and 9 in the second group.

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

Factor out common term 7x-5 by using distributive property.

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

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.

$$35x^{2}+38x-45=0$$

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.

$$x=\frac{-38±\sqrt{38^{2}-4\times 35\left(-45\right)}}{2\times 35}$$

Square 38.

$$x=\frac{-38±\sqrt{1444-4\times 35\left(-45\right)}}{2\times 35}$$

Multiply -4 times 35.

$$x=\frac{-38±\sqrt{1444-140\left(-45\right)}}{2\times 35}$$

Multiply -140 times -45.

$$x=\frac{-38±\sqrt{1444+6300}}{2\times 35}$$

Add 1444 to 6300.

$$x=\frac{-38±\sqrt{7744}}{2\times 35}$$

Take the square root of 7744.

$$x=\frac{-38±88}{2\times 35}$$

Multiply 2 times 35.

$$x=\frac{-38±88}{70}$$

Now solve the equation x=\frac{-38±88}{70} when ± is plus. Add -38 to 88.

$$x=\frac{50}{70}$$

Reduce the fraction \frac{50}{70} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{7}$$

Now solve the equation x=\frac{-38±88}{70} when ± is minus. Subtract 88 from -38.

$$x=-\frac{126}{70}$$

Reduce the fraction \frac{-126}{70} to lowest terms by extracting and canceling out 14.

$$x=-\frac{9}{5}$$

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

$$35x^{2}+38x-45=35\left(x-\frac{5}{7}\right)\left(x-\left(-\frac{9}{5}\right)\right)$$

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

$$35x^{2}+38x-45=35\left(x-\frac{5}{7}\right)\left(x+\frac{9}{5}\right)$$

Subtract \frac{5}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$35x^{2}+38x-45=35\times \frac{7x-5}{7}\left(x+\frac{9}{5}\right)$$

Add \frac{9}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$35x^{2}+38x-45=35\times \frac{7x-5}{7}\times \frac{5x+9}{5}$$

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

$$35x^{2}+38x-45=35\times \frac{\left(7x-5\right)\left(5x+9\right)}{7\times 5}$$

Multiply 7 times 5.

$$35x^{2}+38x-45=35\times \frac{\left(7x-5\right)\left(5x+9\right)}{35}$$

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

$$35x^{2}+38x-45=\left(7x-5\right)\left(5x+9\right)$$

Evaluate

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