Solve 18a^2-51a+15

$$18 a ^ { 2 } - 51 a + 15$$

Factor

$$3\left(2a-5\right)\left(3a-1\right)$$

Solution Steps

Factor out 3.

$$3\left(6a^{2}-17a+5\right)$$

Consider 6a^{2}-17a+5. Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa+5. To find p and q, set up a system to be solved.

$$p+q=-17 pq=6\times 5=30$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -17.

$$p=-15 q=-2$$

Rewrite 6a^{2}-17a+5 as \left(6a^{2}-15a\right)+\left(-2a+5\right).

$$\left(6a^{2}-15a\right)+\left(-2a+5\right)$$

Factor out 3a in the first and -1 in the second group.

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

Factor out common term 2a-5 by using distributive property.

$$\left(2a-5\right)\left(3a-1\right)$$

Rewrite the complete factored expression.

$$3\left(2a-5\right)\left(3a-1\right)$$

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.

$$18a^{2}-51a+15=0$$

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.

$$a=\frac{-\left(-51\right)±\sqrt{\left(-51\right)^{2}-4\times 18\times 15}}{2\times 18}$$

Square -51.

$$a=\frac{-\left(-51\right)±\sqrt{2601-4\times 18\times 15}}{2\times 18}$$

Multiply -4 times 18.

$$a=\frac{-\left(-51\right)±\sqrt{2601-72\times 15}}{2\times 18}$$

Multiply -72 times 15.

$$a=\frac{-\left(-51\right)±\sqrt{2601-1080}}{2\times 18}$$

Add 2601 to -1080.

$$a=\frac{-\left(-51\right)±\sqrt{1521}}{2\times 18}$$

Take the square root of 1521.

$$a=\frac{-\left(-51\right)±39}{2\times 18}$$

The opposite of -51 is 51.

$$a=\frac{51±39}{2\times 18}$$

Multiply 2 times 18.

$$a=\frac{51±39}{36}$$

Now solve the equation a=\frac{51±39}{36} when ± is plus. Add 51 to 39.

$$a=\frac{90}{36}$$

Reduce the fraction \frac{90}{36} to lowest terms by extracting and canceling out 18.

$$a=\frac{5}{2}$$

Now solve the equation a=\frac{51±39}{36} when ± is minus. Subtract 39 from 51.

$$a=\frac{12}{36}$$

Reduce the fraction \frac{12}{36} to lowest terms by extracting and canceling out 12.

$$a=\frac{1}{3}$$

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

$$18a^{2}-51a+15=18\left(a-\frac{5}{2}\right)\left(a-\frac{1}{3}\right)$$

Subtract \frac{5}{2} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18a^{2}-51a+15=18\times \frac{2a-5}{2}\left(a-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18a^{2}-51a+15=18\times \frac{2a-5}{2}\times \frac{3a-1}{3}$$

Multiply \frac{2a-5}{2} times \frac{3a-1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18a^{2}-51a+15=18\times \frac{\left(2a-5\right)\left(3a-1\right)}{2\times 3}$$

Multiply 2 times 3.

$$18a^{2}-51a+15=18\times \frac{\left(2a-5\right)\left(3a-1\right)}{6}$$

Cancel out 6, the greatest common factor in 18 and 6.

$$18a^{2}-51a+15=3\left(2a-5\right)\left(3a-1\right)$$

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 18

$$x ^ 2 -\frac{17}{6}x +\frac{5}{6} = 0$$

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

$$r + s = \frac{17}{6} rs = \frac{5}{6}$$

Two numbers r and s sum up to \frac{17}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{6} = \frac{17}{12}. 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>

$$r = \frac{17}{12} - u s = \frac{17}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{6}

$$(\frac{17}{12} - u) (\frac{17}{12} + u) = \frac{5}{6}$$

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

$$\frac{289}{144} - u^2 = \frac{5}{6}$$

Simplify the expression by subtracting \frac{289}{144} on both sides

$$-u^2 = \frac{5}{6}-\frac{289}{144} = -\frac{169}{144}$$

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

$$u^2 = \frac{169}{144} u = \pm\sqrt{\frac{169}{144}} = \pm \frac{13}{12}$$

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

$$r =\frac{17}{12} - \frac{13}{12} = 0.333 s = \frac{17}{12} + \frac{13}{12} = 2.500$$

Evaluate

$$18a^{2}-51a+15$$