Solve 4x^2+14x+6

$$4 { x }^{ 2 } +14x+6$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(2x^{2}+7x+3\right)$$

Consider 2x^{2}+7x+3. 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.

$$a+b=7 ab=2\times 3=6$$

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 7.

$$a=1 b=6$$

Rewrite 2x^{2}+7x+3 as \left(2x^{2}+x\right)+\left(6x+3\right).

$$\left(2x^{2}+x\right)+\left(6x+3\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$4x^{2}+14x+6=0$$

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.

$$x=\frac{-14±\sqrt{14^{2}-4\times 4\times 6}}{2\times 4}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-14±\sqrt{196-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-14±\sqrt{196-96}}{2\times 4}$$

Add 196 to -96.

$$x=\frac{-14±\sqrt{100}}{2\times 4}$$

Take the square root of 100.

$$x=\frac{-14±10}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-14±10}{8}$$

Now solve the equation x=\frac{-14±10}{8} when ± is plus. Add -14 to 10.

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

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-14±10}{8} when ± is minus. Subtract 10 from -14.

$$x=-\frac{24}{8}$$

Divide -24 by 8.

$$x=-3$$

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 -3 for x_{2}.

$$4x^{2}+14x+6=4\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\left(-3\right)\right)$$

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

$$4x^{2}+14x+6=4\left(x+\frac{1}{2}\right)\left(x+3\right)$$

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

$$4x^{2}+14x+6=4\times \frac{2x+1}{2}\left(x+3\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4x^{2}+14x+6=2\left(2x+1\right)\left(x+3\right)$$

Evaluate

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