Solve 3x^2-42x+135

$$3 { x }^{ 2 } -42x+135$$

Factor

$$3\left(x-9\right)\left(x-5\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{2}-14x+45\right)$$

Consider x^{2}-14x+45. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 45=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

Rewrite x^{2}-14x+45 as \left(x^{2}-9x\right)+\left(-5x+45\right).

$$\left(x^{2}-9x\right)+\left(-5x+45\right)$$

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

$$x\left(x-9\right)-5\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x-5\right)$$

Rewrite the complete factored expression.

$$3\left(x-9\right)\left(x-5\right)$$

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.

$$3x^{2}-42x+135=0$$

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.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\times 3\times 135}}{2\times 3}$$

Square -42.

$$x=\frac{-\left(-42\right)±\sqrt{1764-4\times 3\times 135}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-42\right)±\sqrt{1764-12\times 135}}{2\times 3}$$

Multiply -12 times 135.

$$x=\frac{-\left(-42\right)±\sqrt{1764-1620}}{2\times 3}$$

Add 1764 to -1620.

$$x=\frac{-\left(-42\right)±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{-\left(-42\right)±12}{2\times 3}$$

The opposite of -42 is 42.

$$x=\frac{42±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{42±12}{6}$$

Now solve the equation x=\frac{42±12}{6} when ± is plus. Add 42 to 12.

$$x=\frac{54}{6}$$

Divide 54 by 6.

$$x=9$$

Now solve the equation x=\frac{42±12}{6} when ± is minus. Subtract 12 from 42.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9 for x_{1} and 5 for x_{2}.

$$3x^{2}-42x+135=3\left(x-9\right)\left(x-5\right)$$

Evaluate

$$3\left(x-9\right)\left(x-5\right)$$