Solve 5x^2-7x+2

$$5 { x }^{ 2 } -7x+2$$

Factor

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

Solution Steps

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

$$a+b=-7 ab=5\times 2=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite 5x^{2}-7x+2 as \left(5x^{2}-5x\right)+\left(-2x+2\right).

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

Factor out 5x in the first and -2 in the second group.

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

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

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

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.

$$5x^{2}-7x+2=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5\times 2}}{2\times 5}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 5\times 2}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-7\right)±\sqrt{49-20\times 2}}{2\times 5}$$

Multiply -20 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-40}}{2\times 5}$$

Add 49 to -40.

$$x=\frac{-\left(-7\right)±\sqrt{9}}{2\times 5}$$

Take the square root of 9.

$$x=\frac{-\left(-7\right)±3}{2\times 5}$$

The opposite of -7 is 7.

$$x=\frac{7±3}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{7±3}{10}$$

Now solve the equation x=\frac{7±3}{10} when ± is plus. Add 7 to 3.

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

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{7±3}{10} when ± is minus. Subtract 3 from 7.

$$x=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{5}$$

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{2}{5} for x_{2}.

$$5x^{2}-7x+2=5\left(x-1\right)\left(x-\frac{2}{5}\right)$$

Subtract \frac{2}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-7x+2=5\left(x-1\right)\times \frac{5x-2}{5}$$

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

$$5x^{2}-7x+2=\left(x-1\right)\left(5x-2\right)$$

Evaluate

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