Solve b^2-10b-11

$${ b }^{ 2 } -10b-11$$

Factor

$$\left(b-11\right)\left(b+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-11. To find p and q, set up a system to be solved.

$$p+q=-10 pq=1\left(-11\right)=-11$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$p=-11 q=1$$

Rewrite b^{2}-10b-11 as \left(b^{2}-11b\right)+\left(b-11\right).

$$\left(b^{2}-11b\right)+\left(b-11\right)$$

Factor out b in b^{2}-11b.

$$b\left(b-11\right)+b-11$$

Factor out common term b-11 by using distributive property.

$$\left(b-11\right)\left(b+1\right)$$

Steps Using the Quadratic Formula

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.

$$b^{2}-10b-11=0$$

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.

$$b=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-11\right)}}{2}$$

Square -10.

$$b=\frac{-\left(-10\right)±\sqrt{100-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$b=\frac{-\left(-10\right)±\sqrt{100+44}}{2}$$

Add 100 to 44.

$$b=\frac{-\left(-10\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$b=\frac{-\left(-10\right)±12}{2}$$

The opposite of -10 is 10.

$$b=\frac{10±12}{2}$$

Now solve the equation b=\frac{10±12}{2} when ± is plus. Add 10 to 12.

$$b=\frac{22}{2}$$

Divide 22 by 2.

$$b=11$$

Now solve the equation b=\frac{10±12}{2} when ± is minus. Subtract 12 from 10.

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

Divide -2 by 2.

$$b=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 11 for x_{1} and -1 for x_{2}.

$$b^{2}-10b-11=\left(b-11\right)\left(b-\left(-1\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$b^{2}-10b-11=\left(b-11\right)\left(b+1\right)$$

Evaluate

$$\left(b-11\right)\left(b+1\right)$$