Solve 5x^2-8x+3

$$5 { x }^{ 2 } -8x+3$$

Factor

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

Solution Steps

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

$$a+b=-8 ab=5\times 3=15$$

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 -8.

$$a=-5 b=-3$$

Rewrite 5x^{2}-8x+3 as \left(5x^{2}-5x\right)+\left(-3x+3\right).

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

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

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

Factor out common term x-1 by using distributive property.

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

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.

$$5x^{2}-8x+3=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\times 3}}{2\times 5}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\times 3}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-8\right)±\sqrt{64-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2\times 5}$$

Add 64 to -60.

$$x=\frac{-\left(-8\right)±\sqrt{4}}{2\times 5}$$

Take the square root of 4.

$$x=\frac{-\left(-8\right)±2}{2\times 5}$$

The opposite of -8 is 8.

$$x=\frac{8±2}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{8±2}{10}$$

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

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

Divide 10 by 10.

$$x=1$$

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

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and \frac{3}{5} for x_{2}.

$$5x^{2}-8x+3=5\left(x-1\right)\left(x-\frac{3}{5}\right)$$

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

$$5x^{2}-8x+3=5\left(x-1\right)\times \frac{5x-3}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5x^{2}-8x+3=\left(x-1\right)\left(5x-3\right)$$

Evaluate

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