Solve x^2+26x+133

$${ x }^{ 2 } +26x+133$$

Factor

$$\left(x+7\right)\left(x+19\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+133. To find a and b, set up a system to be solved.

$$a+b=26 ab=1\times 133=133$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 133.

$$1,133 7,19$$

Calculate the sum for each pair.

$$1+133=134 7+19=26$$

The solution is the pair that gives sum 26.

$$a=7 b=19$$

Rewrite x^{2}+26x+133 as \left(x^{2}+7x\right)+\left(19x+133\right).

$$\left(x^{2}+7x\right)+\left(19x+133\right)$$

Factor out x in the first and 19 in the second group.

$$x\left(x+7\right)+19\left(x+7\right)$$

Factor out common term x+7 by using distributive property.

$$\left(x+7\right)\left(x+19\right)$$

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.

$$x^{2}+26x+133=0$$

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.

$$x=\frac{-26±\sqrt{26^{2}-4\times 133}}{2}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\times 133}}{2}$$

Multiply -4 times 133.

$$x=\frac{-26±\sqrt{676-532}}{2}$$

Add 676 to -532.

$$x=\frac{-26±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-26±12}{2}$$

Now solve the equation x=\frac{-26±12}{2} when ± is plus. Add -26 to 12.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

Now solve the equation x=\frac{-26±12}{2} when ± is minus. Subtract 12 from -26.

$$x=-\frac{38}{2}$$

Divide -38 by 2.

$$x=-19$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -7 for x_{1} and -19 for x_{2}.

$$x^{2}+26x+133=\left(x-\left(-7\right)\right)\left(x-\left(-19\right)\right)$$

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

$$x^{2}+26x+133=\left(x+7\right)\left(x+19\right)$$

Evaluate

$$\left(x+7\right)\left(x+19\right)$$