Solve -5u^2-20u+60

$$- 5 u ^ { 2 } - 20 u + 60$$

Factor

$$5\left(2-u\right)\left(u+6\right)$$

Solution Steps

Factor out 5.

$$5\left(-u^{2}-4u+12\right)$$

Consider -u^{2}-4u+12. Factor the expression by grouping. First, the expression needs to be rewritten as -u^{2}+au+bu+12. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-12=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=2 b=-6$$

Rewrite -u^{2}-4u+12 as \left(-u^{2}+2u\right)+\left(-6u+12\right).

$$\left(-u^{2}+2u\right)+\left(-6u+12\right)$$

Factor out u in the first and 6 in the second group.

$$u\left(-u+2\right)+6\left(-u+2\right)$$

Factor out common term -u+2 by using distributive property.

$$\left(-u+2\right)\left(u+6\right)$$

Rewrite the complete factored expression.

$$5\left(-u+2\right)\left(u+6\right)$$

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.

$$-5u^{2}-20u+60=0$$

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.

$$u=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-5\right)\times 60}}{2\left(-5\right)}$$

Square -20.

$$u=\frac{-\left(-20\right)±\sqrt{400-4\left(-5\right)\times 60}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$u=\frac{-\left(-20\right)±\sqrt{400+20\times 60}}{2\left(-5\right)}$$

Multiply 20 times 60.

$$u=\frac{-\left(-20\right)±\sqrt{400+1200}}{2\left(-5\right)}$$

Add 400 to 1200.

$$u=\frac{-\left(-20\right)±\sqrt{1600}}{2\left(-5\right)}$$

Take the square root of 1600.

$$u=\frac{-\left(-20\right)±40}{2\left(-5\right)}$$

The opposite of -20 is 20.

$$u=\frac{20±40}{2\left(-5\right)}$$

Multiply 2 times -5.

$$u=\frac{20±40}{-10}$$

Now solve the equation u=\frac{20±40}{-10} when ± is plus. Add 20 to 40.

$$u=\frac{60}{-10}$$

Divide 60 by -10.

$$u=-6$$

Now solve the equation u=\frac{20±40}{-10} when ± is minus. Subtract 40 from 20.

$$u=-\frac{20}{-10}$$

Divide -20 by -10.

$$u=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6 for x_{1} and 2 for x_{2}.

$$-5u^{2}-20u+60=-5\left(u-\left(-6\right)\right)\left(u-2\right)$$

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

$$-5u^{2}-20u+60=-5\left(u+6\right)\left(u-2\right)$$

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.

$$x ^ 2 +4x -12 = 0$$

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

$$r + s = -4 rs = -12$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. 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>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -12

$$(-2 - u) (-2 + u) = -12$$

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

$$4 - u^2 = -12$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -12-4 = -16$$

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

$$u^2 = 16 u = \pm\sqrt{16} = \pm 4$$

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

$$r =-2 - 4 = -6 s = -2 + 4 = 2$$

Evaluate

$$5\left(2-u\right)\left(u+6\right)$$