Solve (x^2-7x+10)

$$( x ^ { 2 } - 7 x + 10 )$$

Factor

$$\left(x-5\right)\left(x-2\right)$$

Solution Steps

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

$$a+b=-7 ab=1\times 10=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite x^{2}-7x+10 as \left(x^{2}-5x\right)+\left(-2x+10\right).

$$\left(x^{2}-5x\right)+\left(-2x+10\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-5\right)-2\left(x-5\right)$$

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

$$\left(x-5\right)\left(x-2\right)$$

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.

$$x^{2}-7x+10=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 10}}{2}$$

Multiply -4 times 10.

$$x=\frac{-\left(-7\right)±\sqrt{49-40}}{2}$$

Add 49 to -40.

$$x=\frac{-\left(-7\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-7\right)±3}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±3}{2}$$

Now solve the equation x=\frac{7±3}{2} when ± is plus. Add 7 to 3.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{7±3}{2} when ± is minus. Subtract 3 from 7.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

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 2 for x_{2}.

$$x^{2}-7x+10=\left(x-5\right)\left(x-2\right)$$

Evaluate

$$\left(x-5\right)\left(x-2\right)$$