Solve 2x^2+5x+2

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

Factor

$$\left(x+2\right)\left(2x+1\right)$$

Solution Steps

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

$$a+b=5 ab=2\times 2=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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.

$$a=1 b=4$$

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

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

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

$$x\left(2x+1\right)+2\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(x+2\right)$$

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.

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

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 2}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-5±\sqrt{25-16}}{2\times 2}$$

Add 25 to -16.

$$x=\frac{-5±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-5±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±3}{4}$$

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

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

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

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

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

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

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

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

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

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

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

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

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 +\frac{5}{2}x +1 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{5}{2} rs = 1$$

Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{4}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{5}{4} - u s = -\frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = 1$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{25}{16} - u^2 = 1$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = 1-\frac{25}{16} = -\frac{9}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{5}{4} - \frac{3}{4} = -2 s = -\frac{5}{4} + \frac{3}{4} = -0.500$$

Evaluate

$$\left(x+2\right)\left(2x+1\right)$$