Solve (3x^2+16x+5)

$$( 3 x ^ { 2 } + 16 x + 5 )$$

Factor

$$\left(x+5\right)\left(3x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\times 5=15$$

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 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 16.

$$a=1 b=15$$

Rewrite 3x^{2}+16x+5 as \left(3x^{2}+x\right)+\left(15x+5\right).

$$\left(3x^{2}+x\right)+\left(15x+5\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(3x+1\right)+5\left(3x+1\right)$$

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

$$\left(3x+1\right)\left(x+5\right)$$

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.

$$3x^{2}+16x+5=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 3\times 5}}{2\times 3}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-16±\sqrt{256-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$x=\frac{-16±\sqrt{256-60}}{2\times 3}$$

Add 256 to -60.

$$x=\frac{-16±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{-16±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-16±14}{6}$$

Now solve the equation x=\frac{-16±14}{6} when ± is plus. Add -16 to 14.

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

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{-16±14}{6} when ± is minus. Subtract 14 from -16.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

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 -5 for x_{2}.

$$3x^{2}+16x+5=3\left(x-\left(-\frac{1}{3}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$3x^{2}+16x+5=3\left(x+\frac{1}{3}\right)\left(x+5\right)$$

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

$$3x^{2}+16x+5=3\times \frac{3x+1}{3}\left(x+5\right)$$

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

$$3x^{2}+16x+5=\left(3x+1\right)\left(x+5\right)$$

Evaluate

$$\left(x+5\right)\left(3x+1\right)$$