Solve 20z-800+z^2

$$20 z - 800 + z ^ { 2 }$$

Factor

$$\left(z-20\right)\left(z+40\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$z^{2}+20z-800$$

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-800. To find a and b, set up a system to be solved.

$$a+b=20 ab=1\left(-800\right)=-800$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -800.

$$-1,800 -2,400 -4,200 -5,160 -8,100 -10,80 -16,50 -20,40 -25,32$$

Calculate the sum for each pair.

$$-1+800=799 -2+400=398 -4+200=196 -5+160=155 -8+100=92 -10+80=70 -16+50=34 -20+40=20 -25+32=7$$

The solution is the pair that gives sum 20.

$$a=-20 b=40$$

Rewrite z^{2}+20z-800 as \left(z^{2}-20z\right)+\left(40z-800\right).

$$\left(z^{2}-20z\right)+\left(40z-800\right)$$

Factor out z in the first and 40 in the second group.

$$z\left(z-20\right)+40\left(z-20\right)$$

Factor out common term z-20 by using distributive property.

$$\left(z-20\right)\left(z+40\right)$$

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.

$$z^{2}+20z-800=0$$

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.

$$z=\frac{-20±\sqrt{20^{2}-4\left(-800\right)}}{2}$$

Square 20.

$$z=\frac{-20±\sqrt{400-4\left(-800\right)}}{2}$$

Multiply -4 times -800.

$$z=\frac{-20±\sqrt{400+3200}}{2}$$

Add 400 to 3200.

$$z=\frac{-20±\sqrt{3600}}{2}$$

Take the square root of 3600.

$$z=\frac{-20±60}{2}$$

Now solve the equation z=\frac{-20±60}{2} when ± is plus. Add -20 to 60.

$$z=\frac{40}{2}$$

Divide 40 by 2.

$$z=20$$

Now solve the equation z=\frac{-20±60}{2} when ± is minus. Subtract 60 from -20.

$$z=-\frac{80}{2}$$

Divide -80 by 2.

$$z=-40$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 20 for x_{1} and -40 for x_{2}.

$$z^{2}+20z-800=\left(z-20\right)\left(z-\left(-40\right)\right)$$

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

$$z^{2}+20z-800=\left(z-20\right)\left(z+40\right)$$

Evaluate

$$\left(z-20\right)\left(z+40\right)$$