Solve 10x^2-36x+18

$$10 x ^ { 2 } - 36 x + 18$$

Factor

$$2\left(x-3\right)\left(5x-3\right)$$

Solution Steps

Factor out 2.

$$2\left(5x^{2}-18x+9\right)$$

Consider 5x^{2}-18x+9. Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-18 ab=5\times 9=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -18.

$$a=-15 b=-3$$

Rewrite 5x^{2}-18x+9 as \left(5x^{2}-15x\right)+\left(-3x+9\right).

$$\left(5x^{2}-15x\right)+\left(-3x+9\right)$$

Factor out 5x in the first and -3 in the second group.

$$5x\left(x-3\right)-3\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(5x-3\right)$$

Rewrite the complete factored expression.

$$2\left(x-3\right)\left(5x-3\right)$$

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.

$$10x^{2}-36x+18=0$$

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.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 10\times 18}}{2\times 10}$$

Square -36.

$$x=\frac{-\left(-36\right)±\sqrt{1296-4\times 10\times 18}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-36\right)±\sqrt{1296-40\times 18}}{2\times 10}$$

Multiply -40 times 18.

$$x=\frac{-\left(-36\right)±\sqrt{1296-720}}{2\times 10}$$

Add 1296 to -720.

$$x=\frac{-\left(-36\right)±\sqrt{576}}{2\times 10}$$

Take the square root of 576.

$$x=\frac{-\left(-36\right)±24}{2\times 10}$$

The opposite of -36 is 36.

$$x=\frac{36±24}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{36±24}{20}$$

Now solve the equation x=\frac{36±24}{20} when ± is plus. Add 36 to 24.

$$x=\frac{60}{20}$$

Divide 60 by 20.

$$x=3$$

Now solve the equation x=\frac{36±24}{20} when ± is minus. Subtract 24 from 36.

$$x=\frac{12}{20}$$

Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{3}{5} for x_{2}.

$$10x^{2}-36x+18=10\left(x-3\right)\left(x-\frac{3}{5}\right)$$

Subtract \frac{3}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-36x+18=10\left(x-3\right)\times \frac{5x-3}{5}$$

Cancel out 5, the greatest common factor in 10 and 5.

$$10x^{2}-36x+18=2\left(x-3\right)\left(5x-3\right)$$

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 10

$$x ^ 2 -\frac{18}{5}x +\frac{9}{5} = 0$$

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

$$r + s = \frac{18}{5} rs = \frac{9}{5}$$

Two numbers r and s sum up to \frac{18}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{18}{5} = \frac{9}{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>

$$r = \frac{9}{5} - u s = \frac{9}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{5}

$$(\frac{9}{5} - u) (\frac{9}{5} + u) = \frac{9}{5}$$

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

$$\frac{81}{25} - u^2 = \frac{9}{5}$$

Simplify the expression by subtracting \frac{81}{25} on both sides

$$-u^2 = \frac{9}{5}-\frac{81}{25} = -\frac{36}{25}$$

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

$$u^2 = \frac{36}{25} u = \pm\sqrt{\frac{36}{25}} = \pm \frac{6}{5}$$

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

$$r =\frac{9}{5} - \frac{6}{5} = 0.600 s = \frac{9}{5} + \frac{6}{5} = 3$$

Evaluate

$$2\left(x-3\right)\left(5x-3\right)$$