Solve 5x^2-39x+54

$$5 x ^ { 2 } - 39 x + 54$$

Factor

$$\left(x-6\right)\left(5x-9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+54. To find a and b, set up a system to be solved.

$$a+b=-39 ab=5\times 54=270$$

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 270.

$$-1,-270 -2,-135 -3,-90 -5,-54 -6,-45 -9,-30 -10,-27 -15,-18$$

Calculate the sum for each pair.

$$-1-270=-271 -2-135=-137 -3-90=-93 -5-54=-59 -6-45=-51 -9-30=-39 -10-27=-37 -15-18=-33$$

The solution is the pair that gives sum -39.

$$a=-30 b=-9$$

Rewrite 5x^{2}-39x+54 as \left(5x^{2}-30x\right)+\left(-9x+54\right).

$$\left(5x^{2}-30x\right)+\left(-9x+54\right)$$

Factor out 5x in the first and -9 in the second group.

$$5x\left(x-6\right)-9\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(5x-9\right)$$

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.

$$5x^{2}-39x+54=0$$

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.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 5\times 54}}{2\times 5}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\times 5\times 54}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-39\right)±\sqrt{1521-20\times 54}}{2\times 5}$$

Multiply -20 times 54.

$$x=\frac{-\left(-39\right)±\sqrt{1521-1080}}{2\times 5}$$

Add 1521 to -1080.

$$x=\frac{-\left(-39\right)±\sqrt{441}}{2\times 5}$$

Take the square root of 441.

$$x=\frac{-\left(-39\right)±21}{2\times 5}$$

The opposite of -39 is 39.

$$x=\frac{39±21}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{39±21}{10}$$

Now solve the equation x=\frac{39±21}{10} when ± is plus. Add 39 to 21.

$$x=\frac{60}{10}$$

Divide 60 by 10.

$$x=6$$

Now solve the equation x=\frac{39±21}{10} when ± is minus. Subtract 21 from 39.

$$x=\frac{18}{10}$$

Reduce the fraction \frac{18}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and \frac{9}{5} for x_{2}.

$$5x^{2}-39x+54=5\left(x-6\right)\left(x-\frac{9}{5}\right)$$

Subtract \frac{9}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-39x+54=5\left(x-6\right)\times \frac{5x-9}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5x^{2}-39x+54=\left(x-6\right)\left(5x-9\right)$$

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 5

$$x ^ 2 -\frac{39}{5}x +\frac{54}{5} = 0$$

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

$$r + s = \frac{39}{5} rs = \frac{54}{5}$$

Two numbers r and s sum up to \frac{39}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{39}{5} = \frac{39}{10}. 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>

$$r = \frac{39}{10} - u s = \frac{39}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{54}{5}

$$(\frac{39}{10} - u) (\frac{39}{10} + u) = \frac{54}{5}$$

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

$$\frac{1521}{100} - u^2 = \frac{54}{5}$$

Simplify the expression by subtracting \frac{1521}{100} on both sides

$$-u^2 = \frac{54}{5}-\frac{1521}{100} = -\frac{441}{100}$$

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

$$u^2 = \frac{441}{100} u = \pm\sqrt{\frac{441}{100}} = \pm \frac{21}{10}$$

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

$$r =\frac{39}{10} - \frac{21}{10} = 1.800 s = \frac{39}{10} + \frac{21}{10} = 6$$

Evaluate

$$\left(x-6\right)\left(5x-9\right)$$