Solve 6x^2+17x+7

$$6 { x }^{ 2 } +17x+7$$

Factor

$$\left(2x+1\right)\left(3x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=17 ab=6\times 7=42$$

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 42.

$$1,42 2,21 3,14 6,7$$

Calculate the sum for each pair.

$$1+42=43 2+21=23 3+14=17 6+7=13$$

The solution is the pair that gives sum 17.

$$a=3 b=14$$

Rewrite 6x^{2}+17x+7 as \left(6x^{2}+3x\right)+\left(14x+7\right).

$$\left(6x^{2}+3x\right)+\left(14x+7\right)$$

Factor out 3x in the first and 7 in the second group.

$$3x\left(2x+1\right)+7\left(2x+1\right)$$

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

$$\left(2x+1\right)\left(3x+7\right)$$

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.

$$6x^{2}+17x+7=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\times 6\times 7}}{2\times 6}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 6\times 7}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-17±\sqrt{289-24\times 7}}{2\times 6}$$

Multiply -24 times 7.

$$x=\frac{-17±\sqrt{289-168}}{2\times 6}$$

Add 289 to -168.

$$x=\frac{-17±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-17±11}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-17±11}{12}$$

Now solve the equation x=\frac{-17±11}{12} when ± is plus. Add -17 to 11.

$$x=-\frac{6}{12}$$

Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-17±11}{12} when ± is minus. Subtract 11 from -17.

$$x=-\frac{28}{12}$$

Reduce the fraction \frac{-28}{12} to lowest terms by extracting and canceling out 4.

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

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 -\frac{7}{3} for x_{2}.

$$6x^{2}+17x+7=6\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\left(-\frac{7}{3}\right)\right)$$

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

$$6x^{2}+17x+7=6\left(x+\frac{1}{2}\right)\left(x+\frac{7}{3}\right)$$

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

$$6x^{2}+17x+7=6\times \frac{2x+1}{2}\left(x+\frac{7}{3}\right)$$

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

$$6x^{2}+17x+7=6\times \frac{2x+1}{2}\times \frac{3x+7}{3}$$

Multiply \frac{2x+1}{2} times \frac{3x+7}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+17x+7=6\times \frac{\left(2x+1\right)\left(3x+7\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}+17x+7=6\times \frac{\left(2x+1\right)\left(3x+7\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}+17x+7=\left(2x+1\right)\left(3x+7\right)$$

Evaluate

$$\left(2x+1\right)\left(3x+7\right)$$