Solve (8x^2-2x+1)+(3x^2+5x-8)=

$$( 8 x ^ { 2 } - 2 x + 1 ) + ( 3 x ^ { 2 } + 5 x - 8 ) =$$

Evaluate

$$11x^{2}+3x-7$$

Solution Steps

Combine 8x^{2} and 3x^{2} to get 11x^{2}.

$$11x^{2}-2x+1+5x-8$$

Combine -2x and 5x to get 3x.

$$11x^{2}+3x+1-8$$

Subtract 8 from 1 to get -7.

$$11x^{2}+3x-7$$

Factor

$$11\left(x-\frac{-\sqrt{317}-3}{22}\right)\left(x-\frac{\sqrt{317}-3}{22}\right)$$

Steps Using the Quadratic Formula

Combine 8x^{2} and 3x^{2} to get 11x^{2}.

$$factor(11x^{2}-2x+1+5x-8)$$

Combine -2x and 5x to get 3x.

$$factor(11x^{2}+3x+1-8)$$

Subtract 8 from 1 to get -7.

$$factor(11x^{2}+3x-7)$$

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.

$$11x^{2}+3x-7=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\times 11\left(-7\right)}}{2\times 11}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 11\left(-7\right)}}{2\times 11}$$

Multiply -4 times 11.

$$x=\frac{-3±\sqrt{9-44\left(-7\right)}}{2\times 11}$$

Multiply -44 times -7.

$$x=\frac{-3±\sqrt{9+308}}{2\times 11}$$

Add 9 to 308.

$$x=\frac{-3±\sqrt{317}}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{-3±\sqrt{317}}{22}$$

Now solve the equation x=\frac{-3±\sqrt{317}}{22} when ± is plus. Add -3 to \sqrt{317}.

$$x=\frac{\sqrt{317}-3}{22}$$

Now solve the equation x=\frac{-3±\sqrt{317}}{22} when ± is minus. Subtract \sqrt{317} from -3.

$$x=\frac{-\sqrt{317}-3}{22}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{317}}{22} for x_{1} and \frac{-3-\sqrt{317}}{22} for x_{2}.

$$11x^{2}+3x-7=11\left(x-\frac{\sqrt{317}-3}{22}\right)\left(x-\frac{-\sqrt{317}-3}{22}\right)$$