Solve -6x^2-25x-25

$$-6 { x }^{ 2 } -25x-25$$

Factor

$$-\left(2x+5\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -6x^{2}+ax+bx-25. To find a and b, set up a system to be solved.

$$a+b=-25 ab=-6\left(-25\right)=150$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 150.

$$-1,-150 -2,-75 -3,-50 -5,-30 -6,-25 -10,-15$$

Calculate the sum for each pair.

$$-1-150=-151 -2-75=-77 -3-50=-53 -5-30=-35 -6-25=-31 -10-15=-25$$

The solution is the pair that gives sum -25.

$$a=-10 b=-15$$

Rewrite -6x^{2}-25x-25 as \left(-6x^{2}-10x\right)+\left(-15x-25\right).

$$\left(-6x^{2}-10x\right)+\left(-15x-25\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(-3x-5\right)+5\left(-3x-5\right)$$

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

$$\left(-3x-5\right)\left(2x+5\right)$$

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.

$$-6x^{2}-25x-25=0$$

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.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\left(-6\right)\left(-25\right)}}{2\left(-6\right)}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\left(-6\right)\left(-25\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-25\right)±\sqrt{625+24\left(-25\right)}}{2\left(-6\right)}$$

Multiply 24 times -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-600}}{2\left(-6\right)}$$

Add 625 to -600.

$$x=\frac{-\left(-25\right)±\sqrt{25}}{2\left(-6\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-25\right)±5}{2\left(-6\right)}$$

The opposite of -25 is 25.

$$x=\frac{25±5}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{25±5}{-12}$$

Now solve the equation x=\frac{25±5}{-12} when ± is plus. Add 25 to 5.

$$x=\frac{30}{-12}$$

Reduce the fraction \frac{30}{-12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{5}{2}$$

Now solve the equation x=\frac{25±5}{-12} when ± is minus. Subtract 5 from 25.

$$x=\frac{20}{-12}$$

Reduce the fraction \frac{20}{-12} to lowest terms by extracting and canceling out 4.

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

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

$$-6x^{2}-25x-25=-6\left(x-\left(-\frac{5}{2}\right)\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$-6x^{2}-25x-25=-6\left(x+\frac{5}{2}\right)\left(x+\frac{5}{3}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}-25x-25=-6\times \frac{-2x-5}{-2}\left(x+\frac{5}{3}\right)$$

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

$$-6x^{2}-25x-25=-6\times \frac{-2x-5}{-2}\times \frac{-3x-5}{-3}$$

Multiply \frac{-2x-5}{-2} times \frac{-3x-5}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}-25x-25=-6\times \frac{\left(-2x-5\right)\left(-3x-5\right)}{-2\left(-3\right)}$$

Multiply -2 times -3.

$$-6x^{2}-25x-25=-6\times \frac{\left(-2x-5\right)\left(-3x-5\right)}{6}$$

Cancel out 6, the greatest common factor in -6 and 6.

$$-6x^{2}-25x-25=-\left(-2x-5\right)\left(-3x-5\right)$$

Evaluate

$$-6x^{2}-25x-25$$