Solve 3x^2+4x+1

$$3 { x }^{ 2 } +4x+1$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\times 1=3$$

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.

$$a=1 b=3$$

Rewrite 3x^{2}+4x+1 as \left(3x^{2}+x\right)+\left(3x+1\right).

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

Factor out x in 3x^{2}+x.

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

Factor out common term 3x+1 by using distributive property.

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

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.

$$3x^{2}+4x+1=0$$

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.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12}}{2\times 3}$$

Add 16 to -12.

$$x=\frac{-4±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-4±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-4±2}{6}$$

Now solve the equation x=\frac{-4±2}{6} when ± is plus. Add -4 to 2.

$$x=-\frac{2}{6}$$

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

$$x=-\frac{1}{3}$$

Now solve the equation x=\frac{-4±2}{6} when ± is minus. Subtract 2 from -4.

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

Divide -6 by 6.

$$x=-1$$

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

$$3x^{2}+4x+1=3\left(x-\left(-\frac{1}{3}\right)\right)\left(x-\left(-1\right)\right)$$

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

$$3x^{2}+4x+1=3\left(x+\frac{1}{3}\right)\left(x+1\right)$$

Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}+4x+1=3\times \frac{3x+1}{3}\left(x+1\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}+4x+1=\left(3x+1\right)\left(x+1\right)$$

Evaluate

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