Solve 2b^2-b-4-3b^2-b+6

$$2 { b }^{ 2 } -b-4-3 { b }^{ 2 } -b+6$$

Evaluate

$$2-2b-b^{2}$$

Solution Steps

Combine 2b^{2} and -3b^{2} to get -b^{2}.

$$-b^{2}-b-4-b+6$$

Combine -b and -b to get -2b.

$$-b^{2}-2b-4+6$$

Add -4 and 6 to get 2.

$$-b^{2}-2b+2$$

Factor

$$-\left(b-\left(-\sqrt{3}-1\right)\right)\left(b-\left(\sqrt{3}-1\right)\right)$$

Steps Using the Quadratic Formula

Combine 2b^{2} and -3b^{2} to get -b^{2}.

$$factor(-b^{2}-b-4-b+6)$$

Combine -b and -b to get -2b.

$$factor(-b^{2}-2b-4+6)$$

Add -4 and 6 to get 2.

$$factor(-b^{2}-2b+2)$$

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.

$$-b^{2}-2b+2=0$$

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.

$$b=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square -2.

$$b=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$b=\frac{-\left(-2\right)±\sqrt{4+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$b=\frac{-\left(-2\right)±\sqrt{4+8}}{2\left(-1\right)}$$

Add 4 to 8.

$$b=\frac{-\left(-2\right)±\sqrt{12}}{2\left(-1\right)}$$

Take the square root of 12.

$$b=\frac{-\left(-2\right)±2\sqrt{3}}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$b=\frac{2±2\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$b=\frac{2±2\sqrt{3}}{-2}$$

Now solve the equation b=\frac{2±2\sqrt{3}}{-2} when ± is plus. Add 2 to 2\sqrt{3}.

$$b=\frac{2\sqrt{3}+2}{-2}$$

Divide 2+2\sqrt{3} by -2.

$$b=-\left(\sqrt{3}+1\right)$$

Now solve the equation b=\frac{2±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from 2.

$$b=\frac{2-2\sqrt{3}}{-2}$$

Divide 2-2\sqrt{3} by -2.

$$b=\sqrt{3}-1$$

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

$$-b^{2}-2b+2=-\left(b-\left(-\left(\sqrt{3}+1\right)\right)\right)\left(b-\left(\sqrt{3}-1\right)\right)$$