Solve -6a^2-25a-25

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

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -6a^{2}+pa+qa-25. To find p and q, set up a system to be solved.

$$p+q=-25 pq=-6\left(-25\right)=150$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q 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.

$$p=-10 q=-15$$

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

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

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

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

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

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

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.

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

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.

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

Square -25.

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

Multiply -4 times -6.

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

Multiply 24 times -25.

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

Add 625 to -600.

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

Take the square root of 25.

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

The opposite of -25 is 25.

$$a=\frac{25±5}{2\left(-6\right)}$$

Multiply 2 times -6.

$$a=\frac{25±5}{-12}$$

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

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

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

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

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

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

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

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

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}.

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

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

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

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

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

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

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

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

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

Multiply -2 times -3.

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

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

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +\frac{25}{6}x +\frac{25}{6} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{25}{6} rs = \frac{25}{6}$$

Two numbers r and s sum up to -\frac{25}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{25}{6} = -\frac{25}{12}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{25}{12} - u s = -\frac{25}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{6}

$$(-\frac{25}{12} - u) (-\frac{25}{12} + u) = \frac{25}{6}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{625}{144} - u^2 = \frac{25}{6}$$

Simplify the expression by subtracting \frac{625}{144} on both sides

$$-u^2 = \frac{25}{6}-\frac{625}{144} = -\frac{25}{144}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{144} u = \pm\sqrt{\frac{25}{144}} = \pm \frac{5}{12}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{25}{12} - \frac{5}{12} = -2.500 s = -\frac{25}{12} + \frac{5}{12} = -1.667$$

Evaluate

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