Solve 15x^2-26x-57

$$15 { x }^{ 2 } -26x-57$$

Factor

$$\left(x-3\right)\left(15x+19\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx-57. To find a and b, set up a system to be solved.

$$a+b=-26 ab=15\left(-57\right)=-855$$

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 -855.

$$1,-855 3,-285 5,-171 9,-95 15,-57 19,-45$$

Calculate the sum for each pair.

$$1-855=-854 3-285=-282 5-171=-166 9-95=-86 15-57=-42 19-45=-26$$

The solution is the pair that gives sum -26.

$$a=-45 b=19$$

Rewrite 15x^{2}-26x-57 as \left(15x^{2}-45x\right)+\left(19x-57\right).

$$\left(15x^{2}-45x\right)+\left(19x-57\right)$$

Factor out 15x in the first and 19 in the second group.

$$15x\left(x-3\right)+19\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(15x+19\right)$$

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.

$$15x^{2}-26x-57=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 15\left(-57\right)}}{2\times 15}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 15\left(-57\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-26\right)±\sqrt{676-60\left(-57\right)}}{2\times 15}$$

Multiply -60 times -57.

$$x=\frac{-\left(-26\right)±\sqrt{676+3420}}{2\times 15}$$

Add 676 to 3420.

$$x=\frac{-\left(-26\right)±\sqrt{4096}}{2\times 15}$$

Take the square root of 4096.

$$x=\frac{-\left(-26\right)±64}{2\times 15}$$

The opposite of -26 is 26.

$$x=\frac{26±64}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{26±64}{30}$$

Now solve the equation x=\frac{26±64}{30} when ± is plus. Add 26 to 64.

$$x=\frac{90}{30}$$

Divide 90 by 30.

$$x=3$$

Now solve the equation x=\frac{26±64}{30} when ± is minus. Subtract 64 from 26.

$$x=-\frac{38}{30}$$

Reduce the fraction \frac{-38}{30} to lowest terms by extracting and canceling out 2.

$$x=-\frac{19}{15}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -\frac{19}{15} for x_{2}.

$$15x^{2}-26x-57=15\left(x-3\right)\left(x-\left(-\frac{19}{15}\right)\right)$$

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

$$15x^{2}-26x-57=15\left(x-3\right)\left(x+\frac{19}{15}\right)$$

Add \frac{19}{15} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-26x-57=15\left(x-3\right)\times \frac{15x+19}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15x^{2}-26x-57=\left(x-3\right)\left(15x+19\right)$$

Evaluate

$$\left(x-3\right)\left(15x+19\right)$$