Solve 8x^2-10x+3

$$8 { x }^{ 2 } -10x+3$$

Factor

$$\left(2x-1\right)\left(4x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-10 ab=8\times 3=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite 8x^{2}-10x+3 as \left(8x^{2}-6x\right)+\left(-4x+3\right).

$$\left(8x^{2}-6x\right)+\left(-4x+3\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(4x-3\right)-\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(2x-1\right)$$

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.

$$8x^{2}-10x+3=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 8\times 3}}{2\times 8}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 8\times 3}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-10\right)±\sqrt{100-32\times 3}}{2\times 8}$$

Multiply -32 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 8}$$

Add 100 to -96.

$$x=\frac{-\left(-10\right)±\sqrt{4}}{2\times 8}$$

Take the square root of 4.

$$x=\frac{-\left(-10\right)±2}{2\times 8}$$

The opposite of -10 is 10.

$$x=\frac{10±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{10±2}{16}$$

Now solve the equation x=\frac{10±2}{16} when ± is plus. Add 10 to 2.

$$x=\frac{12}{16}$$

Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{4}$$

Now solve the equation x=\frac{10±2}{16} when ± is minus. Subtract 2 from 10.

$$x=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{4} for x_{1} and \frac{1}{2} for x_{2}.

$$8x^{2}-10x+3=8\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{3}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-10x+3=8\times \frac{4x-3}{4}\left(x-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-10x+3=8\times \frac{4x-3}{4}\times \frac{2x-1}{2}$$

Multiply \frac{4x-3}{4} times \frac{2x-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-10x+3=8\times \frac{\left(4x-3\right)\left(2x-1\right)}{4\times 2}$$

Multiply 4 times 2.

$$8x^{2}-10x+3=8\times \frac{\left(4x-3\right)\left(2x-1\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8x^{2}-10x+3=\left(4x-3\right)\left(2x-1\right)$$

Evaluate

$$\left(2x-1\right)\left(4x-3\right)$$