Solve 5w+w^2-50

$$5 w + w ^ { 2 } - 50$$

Factor

$$\left(w-5\right)\left(w+10\right)$$

Solution Steps

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

$$w^{2}+5w-50$$

Factor the expression by grouping. First, the expression needs to be rewritten as w^{2}+aw+bw-50. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-50\right)=-50$$

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 -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 5.

$$a=-5 b=10$$

Rewrite w^{2}+5w-50 as \left(w^{2}-5w\right)+\left(10w-50\right).

$$\left(w^{2}-5w\right)+\left(10w-50\right)$$

Factor out w in the first and 10 in the second group.

$$w\left(w-5\right)+10\left(w-5\right)$$

Factor out common term w-5 by using distributive property.

$$\left(w-5\right)\left(w+10\right)$$

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.

$$w^{2}+5w-50=0$$

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.

$$w=\frac{-5±\sqrt{5^{2}-4\left(-50\right)}}{2}$$

Square 5.

$$w=\frac{-5±\sqrt{25-4\left(-50\right)}}{2}$$

Multiply -4 times -50.

$$w=\frac{-5±\sqrt{25+200}}{2}$$

Add 25 to 200.

$$w=\frac{-5±\sqrt{225}}{2}$$

Take the square root of 225.

$$w=\frac{-5±15}{2}$$

Now solve the equation w=\frac{-5±15}{2} when ± is plus. Add -5 to 15.

$$w=\frac{10}{2}$$

Divide 10 by 2.

$$w=5$$

Now solve the equation w=\frac{-5±15}{2} when ± is minus. Subtract 15 from -5.

$$w=-\frac{20}{2}$$

Divide -20 by 2.

$$w=-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -10 for x_{2}.

$$w^{2}+5w-50=\left(w-5\right)\left(w-\left(-10\right)\right)$$

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

$$w^{2}+5w-50=\left(w-5\right)\left(w+10\right)$$

Evaluate

$$\left(w-5\right)\left(w+10\right)$$