Solve 2x^2+5x+3

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

Factor

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

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.

$$a+b=5 ab=2\times 3=6$$

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 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 5.

$$a=2 b=3$$

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

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

Factor out 2x in the first and 3 in the second group.

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

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

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

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.

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

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 3}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$x=\frac{-5±\sqrt{25-24}}{2\times 2}$$

Add 25 to -24.

$$x=\frac{-5±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-5±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±1}{4}$$

Now solve the equation x=\frac{-5±1}{4} when ± is plus. Add -5 to 1.

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

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-5±1}{4} when ± is minus. Subtract 1 from -5.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{2}$$

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}{2} for x_{2}.

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

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

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

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Evaluate

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