Solve 3k^2+48k+180

$$3 k ^ { 2 } + 48 k + 180$$

Factor

$$3\left(k+6\right)\left(k+10\right)$$

Solution Steps

Factor out 3.

$$3\left(k^{2}+16k+60\right)$$

Consider k^{2}+16k+60. Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk+60. To find a and b, set up a system to be solved.

$$a+b=16 ab=1\times 60=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=6 b=10$$

Rewrite k^{2}+16k+60 as \left(k^{2}+6k\right)+\left(10k+60\right).

$$\left(k^{2}+6k\right)+\left(10k+60\right)$$

Factor out k in the first and 10 in the second group.

$$k\left(k+6\right)+10\left(k+6\right)$$

Factor out common term k+6 by using distributive property.

$$\left(k+6\right)\left(k+10\right)$$

Rewrite the complete factored expression.

$$3\left(k+6\right)\left(k+10\right)$$

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.

$$3k^{2}+48k+180=0$$

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.

$$k=\frac{-48±\sqrt{48^{2}-4\times 3\times 180}}{2\times 3}$$

Square 48.

$$k=\frac{-48±\sqrt{2304-4\times 3\times 180}}{2\times 3}$$

Multiply -4 times 3.

$$k=\frac{-48±\sqrt{2304-12\times 180}}{2\times 3}$$

Multiply -12 times 180.

$$k=\frac{-48±\sqrt{2304-2160}}{2\times 3}$$

Add 2304 to -2160.

$$k=\frac{-48±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$k=\frac{-48±12}{2\times 3}$$

Multiply 2 times 3.

$$k=\frac{-48±12}{6}$$

Now solve the equation k=\frac{-48±12}{6} when ± is plus. Add -48 to 12.

$$k=-\frac{36}{6}$$

Divide -36 by 6.

$$k=-6$$

Now solve the equation k=\frac{-48±12}{6} when ± is minus. Subtract 12 from -48.

$$k=-\frac{60}{6}$$

Divide -60 by 6.

$$k=-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6 for x_{1} and -10 for x_{2}.

$$3k^{2}+48k+180=3\left(k-\left(-6\right)\right)\left(k-\left(-10\right)\right)$$

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

$$3k^{2}+48k+180=3\left(k+6\right)\left(k+10\right)$$

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 3

$$x ^ 2 +16x +60 = 0$$

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

$$r + s = -16 rs = 60$$

Two numbers r and s sum up to -16 exactly when the average of the two numbers is \frac{1}{2}*-16 = -8. 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>

$$r = -8 - u s = -8 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 60

$$(-8 - u) (-8 + u) = 60$$

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

$$64 - u^2 = 60$$

Simplify the expression by subtracting 64 on both sides

$$-u^2 = 60-64 = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =-8 - 2 = -10 s = -8 + 2 = -6$$

Evaluate

$$3\left(k+6\right)\left(k+10\right)$$