Solve (x^2-7x+12)

$$( x ^ { 2 } - 7 x + 12 )$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 12=12$$

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 12.

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

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).

$$\left(x^{2}-4x\right)+\left(-3x+12\right)$$

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

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

Factor out common term x-4 by using distributive property.

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

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.

$$x^{2}-7x+12=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2}$$

Add 49 to -48.

$$x=\frac{-\left(-7\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-7\right)±1}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±1}{2}$$

Now solve the equation x=\frac{7±1}{2} when ± is plus. Add 7 to 1.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{7±1}{2} when ± is minus. Subtract 1 from 7.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and 3 for x_{2}.

$$x^{2}-7x+12=\left(x-4\right)\left(x-3\right)$$

Evaluate

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