Solve (4x^2-7x+3)+(2x^2-2x-5)

$$( 4 x ^ { 2 } - 7 x + 3 ) + ( 2 x ^ { 2 } - 2 x - 5 )$$

Evaluate

$$6x^{2}-9x-2$$

Solution Steps

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$6x^{2}-7x+3-2x-5$$

Combine -7x and -2x to get -9x.

$$6x^{2}-9x+3-5$$

Subtract 5 from 3 to get -2.

$$6x^{2}-9x-2$$

Factor

$$6\left(x-\left(-\frac{\sqrt{129}}{12}+\frac{3}{4}\right)\right)\left(x-\left(\frac{\sqrt{129}}{12}+\frac{3}{4}\right)\right)$$

Steps Using the Quadratic Formula

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$factor(6x^{2}-7x+3-2x-5)$$

Combine -7x and -2x to get -9x.

$$factor(6x^{2}-9x+3-5)$$

Subtract 5 from 3 to get -2.

$$factor(6x^{2}-9x-2)$$

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.

$$6x^{2}-9x-2=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 6\left(-2\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-9\right)±\sqrt{81-24\left(-2\right)}}{2\times 6}$$

Multiply -24 times -2.

$$x=\frac{-\left(-9\right)±\sqrt{81+48}}{2\times 6}$$

Add 81 to 48.

$$x=\frac{-\left(-9\right)±\sqrt{129}}{2\times 6}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{129}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{9±\sqrt{129}}{12}$$

Now solve the equation x=\frac{9±\sqrt{129}}{12} when ± is plus. Add 9 to \sqrt{129}.

$$x=\frac{\sqrt{129}+9}{12}$$

Divide 9+\sqrt{129} by 12.

$$x=\frac{\sqrt{129}}{12}+\frac{3}{4}$$

Now solve the equation x=\frac{9±\sqrt{129}}{12} when ± is minus. Subtract \sqrt{129} from 9.

$$x=\frac{9-\sqrt{129}}{12}$$

Divide 9-\sqrt{129} by 12.

$$x=-\frac{\sqrt{129}}{12}+\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{4}+\frac{\sqrt{129}}{12} for x_{1} and \frac{3}{4}-\frac{\sqrt{129}}{12} for x_{2}.

$$6x^{2}-9x-2=6\left(x-\left(\frac{\sqrt{129}}{12}+\frac{3}{4}\right)\right)\left(x-\left(-\frac{\sqrt{129}}{12}+\frac{3}{4}\right)\right)$$