Solve 3x^2+36x+60

$$3 x ^ { 2 } + 36 x + 60$$

Factor

$$3\left(x+2\right)\left(x+10\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{2}+12x+20\right)$$

Consider x^{2}+12x+20. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 20=20$$

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 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 12.

$$a=2 b=10$$

Rewrite x^{2}+12x+20 as \left(x^{2}+2x\right)+\left(10x+20\right).

$$\left(x^{2}+2x\right)+\left(10x+20\right)$$

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

$$x\left(x+2\right)+10\left(x+2\right)$$

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

$$\left(x+2\right)\left(x+10\right)$$

Rewrite the complete factored expression.

$$3\left(x+2\right)\left(x+10\right)$$

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.

$$3x^{2}+36x+60=0$$

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.

$$x=\frac{-36±\sqrt{36^{2}-4\times 3\times 60}}{2\times 3}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\times 3\times 60}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-36±\sqrt{1296-12\times 60}}{2\times 3}$$

Multiply -12 times 60.

$$x=\frac{-36±\sqrt{1296-720}}{2\times 3}$$

Add 1296 to -720.

$$x=\frac{-36±\sqrt{576}}{2\times 3}$$

Take the square root of 576.

$$x=\frac{-36±24}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-36±24}{6}$$

Now solve the equation x=\frac{-36±24}{6} when ± is plus. Add -36 to 24.

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

Divide -12 by 6.

$$x=-2$$

Now solve the equation x=\frac{-36±24}{6} when ± is minus. Subtract 24 from -36.

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

Divide -60 by 6.

$$x=-10$$

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

$$3x^{2}+36x+60=3\left(x-\left(-2\right)\right)\left(x-\left(-10\right)\right)$$

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

$$3x^{2}+36x+60=3\left(x+2\right)\left(x+10\right)$$

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

$$x ^ 2 +12x +20 = 0$$

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

$$r + s = -12 rs = 20$$

Two numbers r and s sum up to -12 exactly when the average of the two numbers is \frac{1}{2}*-12 = -6. 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>

$$r = -6 - u s = -6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 20

$$(-6 - u) (-6 + u) = 20$$

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

$$36 - u^2 = 20$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = 20-36 = -16$$

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

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

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

$$r =-6 - 4 = -10 s = -6 + 4 = -2$$

Evaluate

$$3\left(x+2\right)\left(x+10\right)$$