Solve f(x)=15x^2+4x-3

$$f ( x ) = 15 x ^ { 2 } + 4 x - 3$$

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.

$$a+b=4 ab=15\left(-3\right)=-45$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -45.

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

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

Rewrite 15x^{2}+4x-3 as \left(15x^{2}-5x\right)+\left(9x-3\right).

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

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

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

Factor out common term 3x-1 by using distributive property.

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

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.

$$15x^{2}+4x-3=0$$

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.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 15\left(-3\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-4±\sqrt{16-60\left(-3\right)}}{2\times 15}$$

Multiply -60 times -3.

$$x=\frac{-4±\sqrt{16+180}}{2\times 15}$$

Add 16 to 180.

$$x=\frac{-4±\sqrt{196}}{2\times 15}$$

Take the square root of 196.

$$x=\frac{-4±14}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-4±14}{30}$$

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

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

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

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

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

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

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

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

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

$$15x^{2}+4x-3=15\left(x-\frac{1}{3}\right)\left(x-\left(-\frac{3}{5}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$15x^{2}+4x-3=15\left(x-\frac{1}{3}\right)\left(x+\frac{3}{5}\right)$$

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

$$15x^{2}+4x-3=15\times \frac{3x-1}{3}\left(x+\frac{3}{5}\right)$$

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

$$15x^{2}+4x-3=15\times \frac{3x-1}{3}\times \frac{5x+3}{5}$$

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

$$15x^{2}+4x-3=15\times \frac{\left(3x-1\right)\left(5x+3\right)}{3\times 5}$$

Multiply 3 times 5.

$$15x^{2}+4x-3=15\times \frac{\left(3x-1\right)\left(5x+3\right)}{15}$$

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

$$15x^{2}+4x-3=\left(3x-1\right)\left(5x+3\right)$$

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 15

$$x ^ 2 +\frac{4}{15}x -\frac{1}{5} = 0$$

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

$$r + s = -\frac{4}{15} rs = -\frac{1}{5}$$

Two numbers r and s sum up to -\frac{4}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{15} = -\frac{2}{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>

$$r = -\frac{2}{15} - u s = -\frac{2}{15} + u$$

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

$$(-\frac{2}{15} - u) (-\frac{2}{15} + u) = -\frac{1}{5}$$

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

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

Simplify the expression by subtracting \frac{4}{225} on both sides

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

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

$$u^2 = \frac{49}{225} u = \pm\sqrt{\frac{49}{225}} = \pm \frac{7}{15}$$

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

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

Evaluate

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