Solve {l}{2x^2-3x+5}{-7x+x^2+1}

$$\left. \begin{array} { l } { 2 x ^ { 2 } - 3 x + 5 } \\ { - 7 x + x ^ { 2 } + 1 } \end{array} \right.$$

Evaluate

$$3x^{2}-10x+6$$

Solution Steps

Combine -3x and -7x to get -10x.

$$2x^{2}-10x+5+x^{2}+1$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}-10x+5+1$$

Add 5 and 1 to get 6.

$$3x^{2}-10x+6$$

Factor

$$3\left(x-\frac{5-\sqrt{7}}{3}\right)\left(x-\frac{\sqrt{7}+5}{3}\right)$$

Steps Using the Quadratic Formula

Combine -3x and -7x to get -10x.

$$factor(2x^{2}-10x+5+x^{2}+1)$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$factor(3x^{2}-10x+5+1)$$

Add 5 and 1 to get 6.

$$factor(3x^{2}-10x+6)$$

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.

$$3x^{2}-10x+6=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 6}}{2\times 3}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 6}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\times 6}}{2\times 3}$$

Multiply -12 times 6.

$$x=\frac{-\left(-10\right)±\sqrt{100-72}}{2\times 3}$$

Add 100 to -72.

$$x=\frac{-\left(-10\right)±\sqrt{28}}{2\times 3}$$

Take the square root of 28.

$$x=\frac{-\left(-10\right)±2\sqrt{7}}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{7}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±2\sqrt{7}}{6}$$

Now solve the equation x=\frac{10±2\sqrt{7}}{6} when ± is plus. Add 10 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+10}{6}$$

Divide 10+2\sqrt{7} by 6.

$$x=\frac{\sqrt{7}+5}{3}$$

Now solve the equation x=\frac{10±2\sqrt{7}}{6} when ± is minus. Subtract 2\sqrt{7} from 10.

$$x=\frac{10-2\sqrt{7}}{6}$$

Divide 10-2\sqrt{7} by 6.

$$x=\frac{5-\sqrt{7}}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{7}}{3} for x_{1} and \frac{5-\sqrt{7}}{3} for x_{2}.

$$3x^{2}-10x+6=3\left(x-\frac{\sqrt{7}+5}{3}\right)\left(x-\frac{5-\sqrt{7}}{3}\right)$$