Solve 3x^2-18x+15

$$3 x ^ { 2 } - 18 x + 15$$

Factor

$$3\left(x-5\right)\left(x-1\right)$$

Solution Steps

Factor out 3.

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

Consider x^{2}-6x+5. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

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

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

Factor out x in the first and -1 in the second group.

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

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

$$\left(x-5\right)\left(x-1\right)$$

Rewrite the complete factored expression.

$$3\left(x-5\right)\left(x-1\right)$$

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.

$$3x^{2}-18x+15=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 15}}{2\times 3}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 15}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-18\right)±\sqrt{324-12\times 15}}{2\times 3}$$

Multiply -12 times 15.

$$x=\frac{-\left(-18\right)±\sqrt{324-180}}{2\times 3}$$

Add 324 to -180.

$$x=\frac{-\left(-18\right)±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{-\left(-18\right)±12}{2\times 3}$$

The opposite of -18 is 18.

$$x=\frac{18±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{18±12}{6}$$

Now solve the equation x=\frac{18±12}{6} when ± is plus. Add 18 to 12.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{18±12}{6} when ± is minus. Subtract 12 from 18.

$$x=\frac{6}{6}$$

Divide 6 by 6.

$$x=1$$

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

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

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 3

$$x ^ 2 -6x +5 = 0$$

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

$$r + s = 6 rs = 5$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(3 - u) (3 + u) = 5$$

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

$$9 - u^2 = 5$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 5-9 = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =3 - 2 = 1 s = 3 + 2 = 5$$

Evaluate

$$3\left(x-5\right)\left(x-1\right)$$