Solve 4x^2-32x+60

$$4 x ^ { 2 } - 32 x + 60$$

Factor

$$4\left(x-5\right)\left(x-3\right)$$

Solution Steps

Factor out 4.

$$4\left(x^{2}-8x+15\right)$$

Consider x^{2}-8x+15. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite x^{2}-8x+15 as \left(x^{2}-5x\right)+\left(-3x+15\right).

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

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

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

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

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

Rewrite the complete factored expression.

$$4\left(x-5\right)\left(x-3\right)$$

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.

$$4x^{2}-32x+60=0$$

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.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 4\times 60}}{2\times 4}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 4\times 60}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-32\right)±\sqrt{1024-16\times 60}}{2\times 4}$$

Multiply -16 times 60.

$$x=\frac{-\left(-32\right)±\sqrt{1024-960}}{2\times 4}$$

Add 1024 to -960.

$$x=\frac{-\left(-32\right)±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{-\left(-32\right)±8}{2\times 4}$$

The opposite of -32 is 32.

$$x=\frac{32±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{32±8}{8}$$

Now solve the equation x=\frac{32±8}{8} when ± is plus. Add 32 to 8.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{32±8}{8} when ± is minus. Subtract 8 from 32.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

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

$$4x^{2}-32x+60=4\left(x-5\right)\left(x-3\right)$$

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.This is achieved by dividing both sides of the equation by 4

$$x ^ 2 -8x +15 = 0$$

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

$$r + s = 8 rs = 15$$

Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 4. 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>

$$r = 4 - u s = 4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 15

$$(4 - u) (4 + u) = 15$$

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

$$16 - u^2 = 15$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 15-16 = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =4 - 1 = 3 s = 4 + 1 = 5$$

Evaluate

$$4\left(x-5\right)\left(x-3\right)$$