Solve 3q^2-143q+1602

$$3 q ^ { 2 } - 143 q + 1602$$

Factor

$$\left(q-18\right)\left(3q-89\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3q^{2}+aq+bq+1602. To find a and b, set up a system to be solved.

$$a+b=-143 ab=3\times 1602=4806$$

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 4806.

$$-1,-4806 -2,-2403 -3,-1602 -6,-801 -9,-534 -18,-267 -27,-178 -54,-89$$

Calculate the sum for each pair.

$$-1-4806=-4807 -2-2403=-2405 -3-1602=-1605 -6-801=-807 -9-534=-543 -18-267=-285 -27-178=-205 -54-89=-143$$

The solution is the pair that gives sum -143.

$$a=-89 b=-54$$

Rewrite 3q^{2}-143q+1602 as \left(3q^{2}-89q\right)+\left(-54q+1602\right).

$$\left(3q^{2}-89q\right)+\left(-54q+1602\right)$$

Factor out q in the first and -18 in the second group.

$$q\left(3q-89\right)-18\left(3q-89\right)$$

Factor out common term 3q-89 by using distributive property.

$$\left(3q-89\right)\left(q-18\right)$$

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.

$$3q^{2}-143q+1602=0$$

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.

$$q=\frac{-\left(-143\right)±\sqrt{\left(-143\right)^{2}-4\times 3\times 1602}}{2\times 3}$$

Square -143.

$$q=\frac{-\left(-143\right)±\sqrt{20449-4\times 3\times 1602}}{2\times 3}$$

Multiply -4 times 3.

$$q=\frac{-\left(-143\right)±\sqrt{20449-12\times 1602}}{2\times 3}$$

Multiply -12 times 1602.

$$q=\frac{-\left(-143\right)±\sqrt{20449-19224}}{2\times 3}$$

Add 20449 to -19224.

$$q=\frac{-\left(-143\right)±\sqrt{1225}}{2\times 3}$$

Take the square root of 1225.

$$q=\frac{-\left(-143\right)±35}{2\times 3}$$

The opposite of -143 is 143.

$$q=\frac{143±35}{2\times 3}$$

Multiply 2 times 3.

$$q=\frac{143±35}{6}$$

Now solve the equation q=\frac{143±35}{6} when ± is plus. Add 143 to 35.

$$q=\frac{178}{6}$$

Reduce the fraction \frac{178}{6} to lowest terms by extracting and canceling out 2.

$$q=\frac{89}{3}$$

Now solve the equation q=\frac{143±35}{6} when ± is minus. Subtract 35 from 143.

$$q=\frac{108}{6}$$

Divide 108 by 6.

$$q=18$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{89}{3} for x_{1} and 18 for x_{2}.

$$3q^{2}-143q+1602=3\left(q-\frac{89}{3}\right)\left(q-18\right)$$

Subtract \frac{89}{3} from q by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3q^{2}-143q+1602=3\times \frac{3q-89}{3}\left(q-18\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3q^{2}-143q+1602=\left(3q-89\right)\left(q-18\right)$$

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

$$x ^ 2 -\frac{143}{3}x +534 = 0$$

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

$$r + s = \frac{143}{3} rs = 534$$

Two numbers r and s sum up to \frac{143}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{143}{3} = \frac{143}{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>

$$r = \frac{143}{6} - u s = \frac{143}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 534

$$(\frac{143}{6} - u) (\frac{143}{6} + u) = 534$$

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

$$\frac{20449}{36} - u^2 = 534$$

Simplify the expression by subtracting \frac{20449}{36} on both sides

$$-u^2 = 534-\frac{20449}{36} = -\frac{1225}{36}$$

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

$$u^2 = \frac{1225}{36} u = \pm\sqrt{\frac{1225}{36}} = \pm \frac{35}{6}$$

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

$$r =\frac{143}{6} - \frac{35}{6} = 18.000 s = \frac{143}{6} + \frac{35}{6} = 29.667$$

Evaluate

$$\left(q-18\right)\left(3q-89\right)$$