Solve 3x+1+2x^2

$$3 x + 1 + 2 x ^ { 2 }$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}+3x+1$$

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\times 1=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite 2x^{2}+3x+1 as \left(2x^{2}+x\right)+\left(2x+1\right).

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

Factor out x in 2x^{2}+x.

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

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

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

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.

$$2x^{2}+3x+1=0$$

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.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8}}{2\times 2}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-3±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±1}{4}$$

Now solve the equation x=\frac{-3±1}{4} when ± is plus. Add -3 to 1.

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

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{-3±1}{4} when ± is minus. Subtract 1 from -3.

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

Divide -4 by 4.

$$x=-1$$

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 -1 for x_{2}.

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

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

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

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

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

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

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

Evaluate

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