Solve 3x^2+30x+48

$$3 x ^ { 2 } + 30 x + 48$$

Factor

$$3\left(x+2\right)\left(x+8\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{2}+10x+16\right)$$

Consider x^{2}+10x+16. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 16=16$$

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 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 10.

$$a=2 b=8$$

Rewrite x^{2}+10x+16 as \left(x^{2}+2x\right)+\left(8x+16\right).

$$\left(x^{2}+2x\right)+\left(8x+16\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x+2\right)+8\left(x+2\right)$$

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

$$\left(x+2\right)\left(x+8\right)$$

Rewrite the complete factored expression.

$$3\left(x+2\right)\left(x+8\right)$$

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.

$$3x^{2}+30x+48=0$$

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.

$$x=\frac{-30±\sqrt{30^{2}-4\times 3\times 48}}{2\times 3}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\times 3\times 48}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-30±\sqrt{900-12\times 48}}{2\times 3}$$

Multiply -12 times 48.

$$x=\frac{-30±\sqrt{900-576}}{2\times 3}$$

Add 900 to -576.

$$x=\frac{-30±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$x=\frac{-30±18}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-30±18}{6}$$

Now solve the equation x=\frac{-30±18}{6} when ± is plus. Add -30 to 18.

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

Divide -12 by 6.

$$x=-2$$

Now solve the equation x=\frac{-30±18}{6} when ± is minus. Subtract 18 from -30.

$$x=-\frac{48}{6}$$

Divide -48 by 6.

$$x=-8$$

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 -8 for x_{2}.

$$3x^{2}+30x+48=3\left(x-\left(-2\right)\right)\left(x-\left(-8\right)\right)$$

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

$$3x^{2}+30x+48=3\left(x+2\right)\left(x+8\right)$$

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

$$x ^ 2 +10x +16 = 0$$

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

$$r + s = -10 rs = 16$$

Two numbers r and s sum up to -10 exactly when the average of the two numbers is \frac{1}{2}*-10 = -5. 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>

$$r = -5 - u s = -5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 16

$$(-5 - u) (-5 + u) = 16$$

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

$$25 - u^2 = 16$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 16-25 = -9$$

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

$$u^2 = 9 u = \pm\sqrt{9} = \pm 3$$

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

$$r =-5 - 3 = -8 s = -5 + 3 = -2$$

Evaluate

$$3\left(x+2\right)\left(x+8\right)$$