Solve 5x^2-28x+36

$$5 x ^ { 2 } - 28 x + 36$$

Factor

$$\left(x-2\right)\left(5x-18\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=-28 ab=5\times 36=180$$

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 180.

$$-1,-180 -2,-90 -3,-60 -4,-45 -5,-36 -6,-30 -9,-20 -10,-18 -12,-15$$

Calculate the sum for each pair.

$$-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27$$

The solution is the pair that gives sum -28.

$$a=-18 b=-10$$

Rewrite 5x^{2}-28x+36 as \left(5x^{2}-18x\right)+\left(-10x+36\right).

$$\left(5x^{2}-18x\right)+\left(-10x+36\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(5x-18\right)-2\left(5x-18\right)$$

Factor out common term 5x-18 by using distributive property.

$$\left(5x-18\right)\left(x-2\right)$$

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.

$$5x^{2}-28x+36=0$$

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.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 5\times 36}}{2\times 5}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 5\times 36}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-28\right)±\sqrt{784-20\times 36}}{2\times 5}$$

Multiply -20 times 36.

$$x=\frac{-\left(-28\right)±\sqrt{784-720}}{2\times 5}$$

Add 784 to -720.

$$x=\frac{-\left(-28\right)±\sqrt{64}}{2\times 5}$$

Take the square root of 64.

$$x=\frac{-\left(-28\right)±8}{2\times 5}$$

The opposite of -28 is 28.

$$x=\frac{28±8}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{28±8}{10}$$

Now solve the equation x=\frac{28±8}{10} when ± is plus. Add 28 to 8.

$$x=\frac{36}{10}$$

Reduce the fraction \frac{36}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{18}{5}$$

Now solve the equation x=\frac{28±8}{10} when ± is minus. Subtract 8 from 28.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{18}{5} for x_{1} and 2 for x_{2}.

$$5x^{2}-28x+36=5\left(x-\frac{18}{5}\right)\left(x-2\right)$$

Subtract \frac{18}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-28x+36=5\times \frac{5x-18}{5}\left(x-2\right)$$

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

$$5x^{2}-28x+36=\left(5x-18\right)\left(x-2\right)$$

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

$$x ^ 2 -\frac{28}{5}x +\frac{36}{5} = 0$$

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

$$r + s = \frac{28}{5} rs = \frac{36}{5}$$

Two numbers r and s sum up to \frac{28}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{28}{5} = \frac{14}{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>

$$r = \frac{14}{5} - u s = \frac{14}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{36}{5}

$$(\frac{14}{5} - u) (\frac{14}{5} + u) = \frac{36}{5}$$

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

$$\frac{196}{25} - u^2 = \frac{36}{5}$$

Simplify the expression by subtracting \frac{196}{25} on both sides

$$-u^2 = \frac{36}{5}-\frac{196}{25} = -\frac{16}{25}$$

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

$$u^2 = \frac{16}{25} u = \pm\sqrt{\frac{16}{25}} = \pm \frac{4}{5}$$

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

$$r =\frac{14}{5} - \frac{4}{5} = 2.000 s = \frac{14}{5} + \frac{4}{5} = 3.600$$

Evaluate

$$\left(x-2\right)\left(5x-18\right)$$