Solve 20-25b+5b^2

$$20 - 25 b + 5 b ^ { 2 }$$

Factor

$$5\left(b-4\right)\left(b-1\right)$$

Solution Steps

Factor out 5.

$$5\left(4-5b+b^{2}\right)$$

Consider 4-5b+b^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$b^{2}-5b+4$$

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb+4. To find p and q, set up a system to be solved.

$$p+q=-5 pq=1\times 4=4$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$p=-4 q=-1$$

Rewrite b^{2}-5b+4 as \left(b^{2}-4b\right)+\left(-b+4\right).

$$\left(b^{2}-4b\right)+\left(-b+4\right)$$

Factor out b in the first and -1 in the second group.

$$b\left(b-4\right)-\left(b-4\right)$$

Factor out common term b-4 by using distributive property.

$$\left(b-4\right)\left(b-1\right)$$

Rewrite the complete factored expression.

$$5\left(b-4\right)\left(b-1\right)$$

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.

$$5b^{2}-25b+20=0$$

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.

$$b=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 5\times 20}}{2\times 5}$$

Square -25.

$$b=\frac{-\left(-25\right)±\sqrt{625-4\times 5\times 20}}{2\times 5}$$

Multiply -4 times 5.

$$b=\frac{-\left(-25\right)±\sqrt{625-20\times 20}}{2\times 5}$$

Multiply -20 times 20.

$$b=\frac{-\left(-25\right)±\sqrt{625-400}}{2\times 5}$$

Add 625 to -400.

$$b=\frac{-\left(-25\right)±\sqrt{225}}{2\times 5}$$

Take the square root of 225.

$$b=\frac{-\left(-25\right)±15}{2\times 5}$$

The opposite of -25 is 25.

$$b=\frac{25±15}{2\times 5}$$

Multiply 2 times 5.

$$b=\frac{25±15}{10}$$

Now solve the equation b=\frac{25±15}{10} when ± is plus. Add 25 to 15.

$$b=\frac{40}{10}$$

Divide 40 by 10.

$$b=4$$

Now solve the equation b=\frac{25±15}{10} when ± is minus. Subtract 15 from 25.

$$b=\frac{10}{10}$$

Divide 10 by 10.

$$b=1$$

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

$$5b^{2}-25b+20=5\left(b-4\right)\left(b-1\right)$$

Evaluate

$$5\left(b-4\right)\left(b-1\right)$$