Solve 3q^2-135q+1350

$$3 q ^ { 2 } - 135 q + 1350$$

Factor

$$3\left(q-30\right)\left(q-15\right)$$

Solution Steps

Factor out 3.

$$3\left(q^{2}-45q+450\right)$$

Consider q^{2}-45q+450. Factor the expression by grouping. First, the expression needs to be rewritten as q^{2}+aq+bq+450. To find a and b, set up a system to be solved.

$$a+b=-45 ab=1\times 450=450$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 450.

$$-1,-450 -2,-225 -3,-150 -5,-90 -6,-75 -9,-50 -10,-45 -15,-30 -18,-25$$

Calculate the sum for each pair.

$$-1-450=-451 -2-225=-227 -3-150=-153 -5-90=-95 -6-75=-81 -9-50=-59 -10-45=-55 -15-30=-45 -18-25=-43$$

The solution is the pair that gives sum -45.

$$a=-30 b=-15$$

Rewrite q^{2}-45q+450 as \left(q^{2}-30q\right)+\left(-15q+450\right).

$$\left(q^{2}-30q\right)+\left(-15q+450\right)$$

Factor out q in the first and -15 in the second group.

$$q\left(q-30\right)-15\left(q-30\right)$$

Factor out common term q-30 by using distributive property.

$$\left(q-30\right)\left(q-15\right)$$

Rewrite the complete factored expression.

$$3\left(q-30\right)\left(q-15\right)$$

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.

$$3q^{2}-135q+1350=0$$

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.

$$q=\frac{-\left(-135\right)±\sqrt{\left(-135\right)^{2}-4\times 3\times 1350}}{2\times 3}$$

Square -135.

$$q=\frac{-\left(-135\right)±\sqrt{18225-4\times 3\times 1350}}{2\times 3}$$

Multiply -4 times 3.

$$q=\frac{-\left(-135\right)±\sqrt{18225-12\times 1350}}{2\times 3}$$

Multiply -12 times 1350.

$$q=\frac{-\left(-135\right)±\sqrt{18225-16200}}{2\times 3}$$

Add 18225 to -16200.

$$q=\frac{-\left(-135\right)±\sqrt{2025}}{2\times 3}$$

Take the square root of 2025.

$$q=\frac{-\left(-135\right)±45}{2\times 3}$$

The opposite of -135 is 135.

$$q=\frac{135±45}{2\times 3}$$

Multiply 2 times 3.

$$q=\frac{135±45}{6}$$

Now solve the equation q=\frac{135±45}{6} when ± is plus. Add 135 to 45.

$$q=\frac{180}{6}$$

Divide 180 by 6.

$$q=30$$

Now solve the equation q=\frac{135±45}{6} when ± is minus. Subtract 45 from 135.

$$q=\frac{90}{6}$$

Divide 90 by 6.

$$q=15$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 30 for x_{1} and 15 for x_{2}.

$$3q^{2}-135q+1350=3\left(q-30\right)\left(q-15\right)$$

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

$$x ^ 2 -45x +450 = 0$$

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

$$r + s = 45 rs = 450$$

Two numbers r and s sum up to 45 exactly when the average of the two numbers is \frac{1}{2}*45 = \frac{45}{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>

$$r = \frac{45}{2} - u s = \frac{45}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 450

$$(\frac{45}{2} - u) (\frac{45}{2} + u) = 450$$

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

$$\frac{2025}{4} - u^2 = 450$$

Simplify the expression by subtracting \frac{2025}{4} on both sides

$$-u^2 = 450-\frac{2025}{4} = -\frac{225}{4}$$

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

$$u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{2}$$

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

$$r =\frac{45}{2} - \frac{15}{2} = 15 s = \frac{45}{2} + \frac{15}{2} = 30$$

Evaluate

$$3\left(q-30\right)\left(q-15\right)$$