Solve 5x^2-145x+600

$$5 { x }^{ 2 } -145x+600$$

Factor

$$5\left(x-24\right)\left(x-5\right)$$

Solution Steps

Factor out 5.

$$5\left(x^{2}-29x+120\right)$$

Consider x^{2}-29x+120. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+120. To find a and b, set up a system to be solved.

$$a+b=-29 ab=1\times 120=120$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -29.

$$a=-24 b=-5$$

Rewrite x^{2}-29x+120 as \left(x^{2}-24x\right)+\left(-5x+120\right).

$$\left(x^{2}-24x\right)+\left(-5x+120\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-24\right)-5\left(x-24\right)$$

Factor out common term x-24 by using distributive property.

$$\left(x-24\right)\left(x-5\right)$$

Rewrite the complete factored expression.

$$5\left(x-24\right)\left(x-5\right)$$

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.

$$5x^{2}-145x+600=0$$

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.

$$x=\frac{-\left(-145\right)±\sqrt{\left(-145\right)^{2}-4\times 5\times 600}}{2\times 5}$$

Square -145.

$$x=\frac{-\left(-145\right)±\sqrt{21025-4\times 5\times 600}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-145\right)±\sqrt{21025-20\times 600}}{2\times 5}$$

Multiply -20 times 600.

$$x=\frac{-\left(-145\right)±\sqrt{21025-12000}}{2\times 5}$$

Add 21025 to -12000.

$$x=\frac{-\left(-145\right)±\sqrt{9025}}{2\times 5}$$

Take the square root of 9025.

$$x=\frac{-\left(-145\right)±95}{2\times 5}$$

The opposite of -145 is 145.

$$x=\frac{145±95}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{145±95}{10}$$

Now solve the equation x=\frac{145±95}{10} when ± is plus. Add 145 to 95.

$$x=\frac{240}{10}$$

Divide 240 by 10.

$$x=24$$

Now solve the equation x=\frac{145±95}{10} when ± is minus. Subtract 95 from 145.

$$x=\frac{50}{10}$$

Divide 50 by 10.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 24 for x_{1} and 5 for x_{2}.

$$5x^{2}-145x+600=5\left(x-24\right)\left(x-5\right)$$

Evaluate

$$5\left(x-24\right)\left(x-5\right)$$