Solve 15+16x-15x^2

$$15 + 16 x - 15 x ^ { 2 }$$

Factor

$$-\left(3x-5\right)\left(5x+3\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-15x^{2}+16x+15$$

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

$$a+b=16 ab=-15\times 15=-225$$

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 -225.

$$-1,225 -3,75 -5,45 -9,25 -15,15$$

Calculate the sum for each pair.

$$-1+225=224 -3+75=72 -5+45=40 -9+25=16 -15+15=0$$

The solution is the pair that gives sum 16.

$$a=25 b=-9$$

Rewrite -15x^{2}+16x+15 as \left(-15x^{2}+25x\right)+\left(-9x+15\right).

$$\left(-15x^{2}+25x\right)+\left(-9x+15\right)$$

Factor out -5x in the first and -3 in the second group.

$$-5x\left(3x-5\right)-3\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(-5x-3\right)$$

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.

$$-15x^{2}+16x+15=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-15\right)\times 15}}{2\left(-15\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-15\right)\times 15}}{2\left(-15\right)}$$

Multiply -4 times -15.

$$x=\frac{-16±\sqrt{256+60\times 15}}{2\left(-15\right)}$$

Multiply 60 times 15.

$$x=\frac{-16±\sqrt{256+900}}{2\left(-15\right)}$$

Add 256 to 900.

$$x=\frac{-16±\sqrt{1156}}{2\left(-15\right)}$$

Take the square root of 1156.

$$x=\frac{-16±34}{2\left(-15\right)}$$

Multiply 2 times -15.

$$x=\frac{-16±34}{-30}$$

Now solve the equation x=\frac{-16±34}{-30} when ± is plus. Add -16 to 34.

$$x=\frac{18}{-30}$$

Reduce the fraction \frac{18}{-30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{3}{5}$$

Now solve the equation x=\frac{-16±34}{-30} when ± is minus. Subtract 34 from -16.

$$x=-\frac{50}{-30}$$

Reduce the fraction \frac{-50}{-30} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{5} for x_{1} and \frac{5}{3} for x_{2}.

$$-15x^{2}+16x+15=-15\left(x-\left(-\frac{3}{5}\right)\right)\left(x-\frac{5}{3}\right)$$

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

$$-15x^{2}+16x+15=-15\left(x+\frac{3}{5}\right)\left(x-\frac{5}{3}\right)$$

Add \frac{3}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-15x^{2}+16x+15=-15\times \frac{-5x-3}{-5}\left(x-\frac{5}{3}\right)$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-15x^{2}+16x+15=-15\times \frac{-5x-3}{-5}\times \frac{-3x+5}{-3}$$

Multiply \frac{-5x-3}{-5} times \frac{-3x+5}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-15x^{2}+16x+15=-15\times \frac{\left(-5x-3\right)\left(-3x+5\right)}{-5\left(-3\right)}$$

Multiply -5 times -3.

$$-15x^{2}+16x+15=-15\times \frac{\left(-5x-3\right)\left(-3x+5\right)}{15}$$

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

$$-15x^{2}+16x+15=-\left(-5x-3\right)\left(-3x+5\right)$$

Evaluate

$$15+16x-15x^{2}$$