Solve 20k^2+110k+120

$$20 k ^ { 2 } + 110 k + 120$$

Factor

$$10\left(k+4\right)\left(2k+3\right)$$

Solution Steps

Factor out 10.

$$10\left(2k^{2}+11k+12\right)$$

Consider 2k^{2}+11k+12. Factor the expression by grouping. First, the expression needs to be rewritten as 2k^{2}+ak+bk+12. To find a and b, set up a system to be solved.

$$a+b=11 ab=2\times 12=24$$

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 2k^{2}+11k+12 as \left(2k^{2}+3k\right)+\left(8k+12\right).

$$\left(2k^{2}+3k\right)+\left(8k+12\right)$$

Factor out k in the first and 4 in the second group.

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

Factor out common term 2k+3 by using distributive property.

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

Rewrite the complete factored expression.

$$10\left(2k+3\right)\left(k+4\right)$$

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.

$$20k^{2}+110k+120=0$$

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.

$$k=\frac{-110±\sqrt{110^{2}-4\times 20\times 120}}{2\times 20}$$

Square 110.

$$k=\frac{-110±\sqrt{12100-4\times 20\times 120}}{2\times 20}$$

Multiply -4 times 20.

$$k=\frac{-110±\sqrt{12100-80\times 120}}{2\times 20}$$

Multiply -80 times 120.

$$k=\frac{-110±\sqrt{12100-9600}}{2\times 20}$$

Add 12100 to -9600.

$$k=\frac{-110±\sqrt{2500}}{2\times 20}$$

Take the square root of 2500.

$$k=\frac{-110±50}{2\times 20}$$

Multiply 2 times 20.

$$k=\frac{-110±50}{40}$$

Now solve the equation k=\frac{-110±50}{40} when ± is plus. Add -110 to 50.

$$k=-\frac{60}{40}$$

Reduce the fraction \frac{-60}{40} to lowest terms by extracting and canceling out 20.

$$k=-\frac{3}{2}$$

Now solve the equation k=\frac{-110±50}{40} when ± is minus. Subtract 50 from -110.

$$k=-\frac{160}{40}$$

Divide -160 by 40.

$$k=-4$$

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

$$20k^{2}+110k+120=20\left(k-\left(-\frac{3}{2}\right)\right)\left(k-\left(-4\right)\right)$$

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

$$20k^{2}+110k+120=20\left(k+\frac{3}{2}\right)\left(k+4\right)$$

Add \frac{3}{2} to k by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20k^{2}+110k+120=20\times \frac{2k+3}{2}\left(k+4\right)$$

Cancel out 2, the greatest common factor in 20 and 2.

$$20k^{2}+110k+120=10\left(2k+3\right)\left(k+4\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 20

$$x ^ 2 +\frac{11}{2}x +6 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{11}{2} rs = 6$$

Two numbers r and s sum up to -\frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{2} = -\frac{11}{4}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{11}{4} - u s = -\frac{11}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 6

$$(-\frac{11}{4} - u) (-\frac{11}{4} + u) = 6$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{121}{16} - u^2 = 6$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = 6-\frac{121}{16} = -\frac{25}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{11}{4} - \frac{5}{4} = -4 s = -\frac{11}{4} + \frac{5}{4} = -1.500$$

Evaluate

$$10\left(k+4\right)\left(2k+3\right)$$