Solve (2a^2-5a+2)

$$( 2 a ^ { 2 } - 5 a + 2 )$$

Factor

$$\left(a-2\right)\left(2a-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2a^{2}+pa+qa+2. To find p and q, set up a system to be solved.

$$p+q=-5 pq=2\times 2=4$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$p=-4 q=-1$$

Rewrite 2a^{2}-5a+2 as \left(2a^{2}-4a\right)+\left(-a+2\right).

$$\left(2a^{2}-4a\right)+\left(-a+2\right)$$

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

$$2a\left(a-2\right)-\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(2a-1\right)$$

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.

$$2a^{2}-5a+2=0$$

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.

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

Square -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-\left(-5\right)±\sqrt{25-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$a=\frac{-\left(-5\right)±\sqrt{25-16}}{2\times 2}$$

Add 25 to -16.

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

Take the square root of 9.

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

The opposite of -5 is 5.

$$a=\frac{5±3}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{5±3}{4}$$

Now solve the equation a=\frac{5±3}{4} when ± is plus. Add 5 to 3.

$$a=\frac{8}{4}$$

Divide 8 by 4.

$$a=2$$

Now solve the equation a=\frac{5±3}{4} when ± is minus. Subtract 3 from 5.

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

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

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

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

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

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

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

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

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

Evaluate

$$\left(a-2\right)\left(2a-1\right)$$