Solve 4x^2-40x+96

$$4 { x }^{ 2 } -40x+96$$

Factor

$$4\left(x-6\right)\left(x-4\right)$$

Solution Steps

Factor out 4.

$$4\left(x^{2}-10x+24\right)$$

Consider x^{2}-10x+24. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite x^{2}-10x+24 as \left(x^{2}-6x\right)+\left(-4x+24\right).

$$\left(x^{2}-6x\right)+\left(-4x+24\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-6\right)-4\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-4\right)$$

Rewrite the complete factored expression.

$$4\left(x-6\right)\left(x-4\right)$$

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.

$$4x^{2}-40x+96=0$$

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.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 4\times 96}}{2\times 4}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 4\times 96}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-40\right)±\sqrt{1600-16\times 96}}{2\times 4}$$

Multiply -16 times 96.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1536}}{2\times 4}$$

Add 1600 to -1536.

$$x=\frac{-\left(-40\right)±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{-\left(-40\right)±8}{2\times 4}$$

The opposite of -40 is 40.

$$x=\frac{40±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{40±8}{8}$$

Now solve the equation x=\frac{40±8}{8} when ± is plus. Add 40 to 8.

$$x=\frac{48}{8}$$

Divide 48 by 8.

$$x=6$$

Now solve the equation x=\frac{40±8}{8} when ± is minus. Subtract 8 from 40.

$$x=\frac{32}{8}$$

Divide 32 by 8.

$$x=4$$

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 4 for x_{2}.

$$4x^{2}-40x+96=4\left(x-6\right)\left(x-4\right)$$

Evaluate

$$4\left(x-6\right)\left(x-4\right)$$