Solve 3x^2-17x+20

$$3 { x }^{ 2 } -17x+20$$

Factor

$$\left(x-4\right)\left(3x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-17 ab=3\times 20=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -17.

$$a=-12 b=-5$$

Rewrite 3x^{2}-17x+20 as \left(3x^{2}-12x\right)+\left(-5x+20\right).

$$\left(3x^{2}-12x\right)+\left(-5x+20\right)$$

Factor out 3x in the first and -5 in the second group.

$$3x\left(x-4\right)-5\left(x-4\right)$$

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

$$\left(x-4\right)\left(3x-5\right)$$

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.

$$3x^{2}-17x+20=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 3\times 20}}{2\times 3}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 3\times 20}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-17\right)±\sqrt{289-12\times 20}}{2\times 3}$$

Multiply -12 times 20.

$$x=\frac{-\left(-17\right)±\sqrt{289-240}}{2\times 3}$$

Add 289 to -240.

$$x=\frac{-\left(-17\right)±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$x=\frac{-\left(-17\right)±7}{2\times 3}$$

The opposite of -17 is 17.

$$x=\frac{17±7}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{17±7}{6}$$

Now solve the equation x=\frac{17±7}{6} when ± is plus. Add 17 to 7.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{17±7}{6} when ± is minus. Subtract 7 from 17.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

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 \frac{5}{3} for x_{2}.

$$3x^{2}-17x+20=3\left(x-4\right)\left(x-\frac{5}{3}\right)$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-17x+20=3\left(x-4\right)\times \frac{3x-5}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}-17x+20=\left(x-4\right)\left(3x-5\right)$$

Evaluate

$$\left(x-4\right)\left(3x-5\right)$$