Solve 3x^2+16x+21

$$3 { x }^{ 2 } +16x+21$$

Factor

$$\left(x+3\right)\left(3x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\times 21=63$$

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 63.

$$1,63 3,21 7,9$$

Calculate the sum for each pair.

$$1+63=64 3+21=24 7+9=16$$

The solution is the pair that gives sum 16.

$$a=7 b=9$$

Rewrite 3x^{2}+16x+21 as \left(3x^{2}+7x\right)+\left(9x+21\right).

$$\left(3x^{2}+7x\right)+\left(9x+21\right)$$

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

$$x\left(3x+7\right)+3\left(3x+7\right)$$

Factor out common term 3x+7 by using distributive property.

$$\left(3x+7\right)\left(x+3\right)$$

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.

$$3x^{2}+16x+21=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 3\times 21}}{2\times 3}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 3\times 21}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-16±\sqrt{256-12\times 21}}{2\times 3}$$

Multiply -12 times 21.

$$x=\frac{-16±\sqrt{256-252}}{2\times 3}$$

Add 256 to -252.

$$x=\frac{-16±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-16±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-16±2}{6}$$

Now solve the equation x=\frac{-16±2}{6} when ± is plus. Add -16 to 2.

$$x=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{3}$$

Now solve the equation x=\frac{-16±2}{6} when ± is minus. Subtract 2 from -16.

$$x=-\frac{18}{6}$$

Divide -18 by 6.

$$x=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{3} for x_{1} and -3 for x_{2}.

$$3x^{2}+16x+21=3\left(x-\left(-\frac{7}{3}\right)\right)\left(x-\left(-3\right)\right)$$

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

$$3x^{2}+16x+21=3\left(x+\frac{7}{3}\right)\left(x+3\right)$$

Add \frac{7}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}+16x+21=3\times \frac{3x+7}{3}\left(x+3\right)$$

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

$$3x^{2}+16x+21=\left(3x+7\right)\left(x+3\right)$$

Evaluate

$$\left(x+3\right)\left(3x+7\right)$$