Solve (2x^2-7x+3

$$(2 { x }^{ 2 } -7x+3$$

Factor

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

Solution Steps

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.

$$a+b=-7 ab=2\times 3=6$$

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 -7.

$$a=-6 b=-1$$

Rewrite 2x^{2}-7x+3 as \left(2x^{2}-6x\right)+\left(-x+3\right).

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

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

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

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

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

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.

$$2x^{2}-7x+3=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 3}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2\times 2}$$

Add 49 to -24.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

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

The opposite of -7 is 7.

$$x=\frac{7±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±5}{4}$$

Now solve the equation x=\frac{7±5}{4} when ± is plus. Add 7 to 5.

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

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{7±5}{4} when ± is minus. Subtract 5 from 7.

$$x=\frac{2}{4}$$

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

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

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

$$2x^{2}-7x+3=2\left(x-3\right)\left(x-\frac{1}{2}\right)$$

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

$$2x^{2}-7x+3=2\left(x-3\right)\times \frac{2x-1}{2}$$

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

$$2x^{2}-7x+3=\left(x-3\right)\left(2x-1\right)$$

Evaluate

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