Solve 7x^2-3x+x^2-3-5x-1=

$$7 x ^ { 2 } - 3 x + x ^ { 2 } - 3 - 5 x - 1 =$$

Evaluate

$$8x^{2}-8x-4$$

Solution Steps

Combine 7x^{2} and x^{2} to get 8x^{2}.

$$8x^{2}-3x-3-5x-1$$

Combine -3x and -5x to get -8x.

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

Subtract 1 from -3 to get -4.

$$8x^{2}-8x-4$$

Factor

$$8\left(x-\frac{1-\sqrt{3}}{2}\right)\left(x-\frac{\sqrt{3}+1}{2}\right)$$

Steps Using the Quadratic Formula

Combine 7x^{2} and x^{2} to get 8x^{2}.

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

Combine -3x and -5x to get -8x.

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

Subtract 1 from -3 to get -4.

$$factor(8x^{2}-8x-4)$$

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.

$$8x^{2}-8x-4=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 8\left(-4\right)}}{2\times 8}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 8\left(-4\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-8\right)±\sqrt{64-32\left(-4\right)}}{2\times 8}$$

Multiply -32 times -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+128}}{2\times 8}$$

Add 64 to 128.

$$x=\frac{-\left(-8\right)±\sqrt{192}}{2\times 8}$$

Take the square root of 192.

$$x=\frac{-\left(-8\right)±8\sqrt{3}}{2\times 8}$$

The opposite of -8 is 8.

$$x=\frac{8±8\sqrt{3}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{8±8\sqrt{3}}{16}$$

Now solve the equation x=\frac{8±8\sqrt{3}}{16} when ± is plus. Add 8 to 8\sqrt{3}.

$$x=\frac{8\sqrt{3}+8}{16}$$

Divide 8+8\sqrt{3} by 16.

$$x=\frac{\sqrt{3}+1}{2}$$

Now solve the equation x=\frac{8±8\sqrt{3}}{16} when ± is minus. Subtract 8\sqrt{3} from 8.

$$x=\frac{8-8\sqrt{3}}{16}$$

Divide 8-8\sqrt{3} by 16.

$$x=\frac{1-\sqrt{3}}{2}$$

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

$$8x^{2}-8x-4=8\left(x-\frac{\sqrt{3}+1}{2}\right)\left(x-\frac{1-\sqrt{3}}{2}\right)$$