Solve 6x^2-32x+10

$$6 { x }^{ 2 } -32x+10$$

Factor

$$2\left(x-5\right)\left(3x-1\right)$$

Solution Steps

Factor out 2.

$$2\left(3x^{2}-16x+5\right)$$

Consider 3x^{2}-16x+5. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-16 ab=3\times 5=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -16.

$$a=-15 b=-1$$

Rewrite 3x^{2}-16x+5 as \left(3x^{2}-15x\right)+\left(-x+5\right).

$$\left(3x^{2}-15x\right)+\left(-x+5\right)$$

Factor out 3x in the first and -1 in the second group.

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

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

$$\left(x-5\right)\left(3x-1\right)$$

Rewrite the complete factored expression.

$$2\left(x-5\right)\left(3x-1\right)$$

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.

$$6x^{2}-32x+10=0$$

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.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 6\times 10}}{2\times 6}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 6\times 10}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-32\right)±\sqrt{1024-24\times 10}}{2\times 6}$$

Multiply -24 times 10.

$$x=\frac{-\left(-32\right)±\sqrt{1024-240}}{2\times 6}$$

Add 1024 to -240.

$$x=\frac{-\left(-32\right)±\sqrt{784}}{2\times 6}$$

Take the square root of 784.

$$x=\frac{-\left(-32\right)±28}{2\times 6}$$

The opposite of -32 is 32.

$$x=\frac{32±28}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{32±28}{12}$$

Now solve the equation x=\frac{32±28}{12} when ± is plus. Add 32 to 28.

$$x=\frac{60}{12}$$

Divide 60 by 12.

$$x=5$$

Now solve the equation x=\frac{32±28}{12} when ± is minus. Subtract 28 from 32.

$$x=\frac{4}{12}$$

Reduce the fraction \frac{4}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{3}$$

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 \frac{1}{3} for x_{2}.

$$6x^{2}-32x+10=6\left(x-5\right)\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-32x+10=6\left(x-5\right)\times \frac{3x-1}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}-32x+10=2\left(x-5\right)\left(3x-1\right)$$

Evaluate

$$2\left(x-5\right)\left(3x-1\right)$$