Solve 15x^2-14x+3

$$15 x ^ { 2 } - 14 x + 3$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-14 ab=15\times 3=45$$

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 -14.

$$a=-9 b=-5$$

Rewrite 15x^{2}-14x+3 as \left(15x^{2}-9x\right)+\left(-5x+3\right).

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

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

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

Factor out common term 5x-3 by using distributive property.

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

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.

$$15x^{2}-14x+3=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 15\times 3}}{2\times 15}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 15\times 3}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-14\right)±\sqrt{196-60\times 3}}{2\times 15}$$

Multiply -60 times 3.

$$x=\frac{-\left(-14\right)±\sqrt{196-180}}{2\times 15}$$

Add 196 to -180.

$$x=\frac{-\left(-14\right)±\sqrt{16}}{2\times 15}$$

Take the square root of 16.

$$x=\frac{-\left(-14\right)±4}{2\times 15}$$

The opposite of -14 is 14.

$$x=\frac{14±4}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{14±4}{30}$$

Now solve the equation x=\frac{14±4}{30} when ± is plus. Add 14 to 4.

$$x=\frac{18}{30}$$

Reduce the fraction \frac{18}{30} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{14±4}{30} when ± is minus. Subtract 4 from 14.

$$x=\frac{10}{30}$$

Reduce the fraction \frac{10}{30} to lowest terms by extracting and canceling out 10.

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

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

$$15x^{2}-14x+3=15\left(x-\frac{3}{5}\right)\left(x-\frac{1}{3}\right)$$

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

$$15x^{2}-14x+3=15\times \frac{5x-3}{5}\left(x-\frac{1}{3}\right)$$

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

$$15x^{2}-14x+3=15\times \frac{5x-3}{5}\times \frac{3x-1}{3}$$

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

$$15x^{2}-14x+3=15\times \frac{\left(5x-3\right)\left(3x-1\right)}{5\times 3}$$

Multiply 5 times 3.

$$15x^{2}-14x+3=15\times \frac{\left(5x-3\right)\left(3x-1\right)}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15x^{2}-14x+3=\left(5x-3\right)\left(3x-1\right)$$

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 15

$$x ^ 2 -\frac{14}{15}x +\frac{1}{5} = 0$$

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

$$r + s = \frac{14}{15} rs = \frac{1}{5}$$

Two numbers r and s sum up to \frac{14}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{14}{15} = \frac{7}{15}. 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>

$$r = \frac{7}{15} - u s = \frac{7}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{5}

$$(\frac{7}{15} - u) (\frac{7}{15} + u) = \frac{1}{5}$$

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

$$\frac{49}{225} - u^2 = \frac{1}{5}$$

Simplify the expression by subtracting \frac{49}{225} on both sides

$$-u^2 = \frac{1}{5}-\frac{49}{225} = -\frac{4}{225}$$

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

$$u^2 = \frac{4}{225} u = \pm\sqrt{\frac{4}{225}} = \pm \frac{2}{15}$$

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

$$r =\frac{7}{15} - \frac{2}{15} = 0.333 s = \frac{7}{15} + \frac{2}{15} = 0.600$$

Evaluate

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