Solve (4k^2+11k+6)

$$( 4 k ^ { 2 } + 11 k + 6 )$$

Factor

$$\left(k+2\right)\left(4k+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4k^{2}+ak+bk+6. To find a and b, set up a system to be solved.

$$a+b=11 ab=4\times 6=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=3 b=8$$

Rewrite 4k^{2}+11k+6 as \left(4k^{2}+3k\right)+\left(8k+6\right).

$$\left(4k^{2}+3k\right)+\left(8k+6\right)$$

Factor out k in the first and 2 in the second group.

$$k\left(4k+3\right)+2\left(4k+3\right)$$

Factor out common term 4k+3 by using distributive property.

$$\left(4k+3\right)\left(k+2\right)$$

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.

$$4k^{2}+11k+6=0$$

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.

$$k=\frac{-11±\sqrt{11^{2}-4\times 4\times 6}}{2\times 4}$$

Square 11.

$$k=\frac{-11±\sqrt{121-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$k=\frac{-11±\sqrt{121-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$k=\frac{-11±\sqrt{121-96}}{2\times 4}$$

Add 121 to -96.

$$k=\frac{-11±\sqrt{25}}{2\times 4}$$

Take the square root of 25.

$$k=\frac{-11±5}{2\times 4}$$

Multiply 2 times 4.

$$k=\frac{-11±5}{8}$$

Now solve the equation k=\frac{-11±5}{8} when ± is plus. Add -11 to 5.

$$k=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$k=-\frac{3}{4}$$

Now solve the equation k=\frac{-11±5}{8} when ± is minus. Subtract 5 from -11.

$$k=-\frac{16}{8}$$

Divide -16 by 8.

$$k=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{4} for x_{1} and -2 for x_{2}.

$$4k^{2}+11k+6=4\left(k-\left(-\frac{3}{4}\right)\right)\left(k-\left(-2\right)\right)$$

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

$$4k^{2}+11k+6=4\left(k+\frac{3}{4}\right)\left(k+2\right)$$

Add \frac{3}{4} to k by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4k^{2}+11k+6=4\times \frac{4k+3}{4}\left(k+2\right)$$

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

$$4k^{2}+11k+6=\left(4k+3\right)\left(k+2\right)$$

Evaluate

$$\left(k+2\right)\left(4k+3\right)$$