Solve 4x^2-10x+6

$$4 { x }^{ 2 } -10x+6$$

Factor

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

Solution Steps

Factor out 2.

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

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

$$a+b=-5 ab=2\times 3=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

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

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

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

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

Factor out common term 2x-3 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$4x^{2}-10x+6=0$$

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.

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

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-10\right)±\sqrt{100-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 4}$$

Add 100 to -96.

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

Take the square root of 4.

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

The opposite of -10 is 10.

$$x=\frac{10±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{10±2}{8}$$

Now solve the equation x=\frac{10±2}{8} when ± is plus. Add 10 to 2.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{10±2}{8} when ± is minus. Subtract 2 from 10.

$$x=\frac{8}{8}$$

Divide 8 by 8.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and 1 for x_{2}.

$$4x^{2}-10x+6=4\left(x-\frac{3}{2}\right)\left(x-1\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-10x+6=4\times \frac{2x-3}{2}\left(x-1\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4x^{2}-10x+6=2\left(2x-3\right)\left(x-1\right)$$

Evaluate

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