Solve 3u^2+51u+90

$$3 u ^ { 2 } + 51 u + 90$$

Factor

$$3\left(u+2\right)\left(u+15\right)$$

Solution Steps

Factor out 3.

$$3\left(u^{2}+17u+30\right)$$

Consider u^{2}+17u+30. Factor the expression by grouping. First, the expression needs to be rewritten as u^{2}+au+bu+30. To find a and b, set up a system to be solved.

$$a+b=17 ab=1\times 30=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 17.

$$a=2 b=15$$

Rewrite u^{2}+17u+30 as \left(u^{2}+2u\right)+\left(15u+30\right).

$$\left(u^{2}+2u\right)+\left(15u+30\right)$$

Factor out u in the first and 15 in the second group.

$$u\left(u+2\right)+15\left(u+2\right)$$

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

$$\left(u+2\right)\left(u+15\right)$$

Rewrite the complete factored expression.

$$3\left(u+2\right)\left(u+15\right)$$

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.

$$3u^{2}+51u+90=0$$

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.

$$u=\frac{-51±\sqrt{51^{2}-4\times 3\times 90}}{2\times 3}$$

Square 51.

$$u=\frac{-51±\sqrt{2601-4\times 3\times 90}}{2\times 3}$$

Multiply -4 times 3.

$$u=\frac{-51±\sqrt{2601-12\times 90}}{2\times 3}$$

Multiply -12 times 90.

$$u=\frac{-51±\sqrt{2601-1080}}{2\times 3}$$

Add 2601 to -1080.

$$u=\frac{-51±\sqrt{1521}}{2\times 3}$$

Take the square root of 1521.

$$u=\frac{-51±39}{2\times 3}$$

Multiply 2 times 3.

$$u=\frac{-51±39}{6}$$

Now solve the equation u=\frac{-51±39}{6} when ± is plus. Add -51 to 39.

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

Divide -12 by 6.

$$u=-2$$

Now solve the equation u=\frac{-51±39}{6} when ± is minus. Subtract 39 from -51.

$$u=-\frac{90}{6}$$

Divide -90 by 6.

$$u=-15$$

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 -15 for x_{2}.

$$3u^{2}+51u+90=3\left(u-\left(-2\right)\right)\left(u-\left(-15\right)\right)$$

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

$$3u^{2}+51u+90=3\left(u+2\right)\left(u+15\right)$$

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

$$x ^ 2 +17x +30 = 0$$

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

$$r + s = -17 rs = 30$$

Two numbers r and s sum up to -17 exactly when the average of the two numbers is \frac{1}{2}*-17 = -\frac{17}{2}. 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>

$$r = -\frac{17}{2} - u s = -\frac{17}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 30

$$(-\frac{17}{2} - u) (-\frac{17}{2} + u) = 30$$

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

$$\frac{289}{4} - u^2 = 30$$

Simplify the expression by subtracting \frac{289}{4} on both sides

$$-u^2 = 30-\frac{289}{4} = -\frac{169}{4}$$

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

$$u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}$$

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

$$r =-\frac{17}{2} - \frac{13}{2} = -15 s = -\frac{17}{2} + \frac{13}{2} = -2$$

Evaluate

$$3\left(u+2\right)\left(u+15\right)$$