Solve 4c^2+15c-4

$$4 c ^ { 2 } + 15 c - 4$$

Factor

$$\left(4c-1\right)\left(c+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4c^{2}+ac+bc-4. To find a and b, set up a system to be solved.

$$a+b=15 ab=4\left(-4\right)=-16$$

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 -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 15.

$$a=-1 b=16$$

Rewrite 4c^{2}+15c-4 as \left(4c^{2}-c\right)+\left(16c-4\right).

$$\left(4c^{2}-c\right)+\left(16c-4\right)$$

Factor out c in the first and 4 in the second group.

$$c\left(4c-1\right)+4\left(4c-1\right)$$

Factor out common term 4c-1 by using distributive property.

$$\left(4c-1\right)\left(c+4\right)$$

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.

$$4c^{2}+15c-4=0$$

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.

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

Square 15.

$$c=\frac{-15±\sqrt{225-4\times 4\left(-4\right)}}{2\times 4}$$

Multiply -4 times 4.

$$c=\frac{-15±\sqrt{225-16\left(-4\right)}}{2\times 4}$$

Multiply -16 times -4.

$$c=\frac{-15±\sqrt{225+64}}{2\times 4}$$

Add 225 to 64.

$$c=\frac{-15±\sqrt{289}}{2\times 4}$$

Take the square root of 289.

$$c=\frac{-15±17}{2\times 4}$$

Multiply 2 times 4.

$$c=\frac{-15±17}{8}$$

Now solve the equation c=\frac{-15±17}{8} when ± is plus. Add -15 to 17.

$$c=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$c=\frac{1}{4}$$

Now solve the equation c=\frac{-15±17}{8} when ± is minus. Subtract 17 from -15.

$$c=-\frac{32}{8}$$

Divide -32 by 8.

$$c=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4} for x_{1} and -4 for x_{2}.

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

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

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

Subtract \frac{1}{4} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 4, the greatest common factor in 4 and 4.

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

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 4

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

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

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

Two numbers r and s sum up to -\frac{15}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{15}{4} = -\frac{15}{8}. 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>

$$r = -\frac{15}{8} - u s = -\frac{15}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(-\frac{15}{8} - u) (-\frac{15}{8} + u) = -1$$

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

$$\frac{225}{64} - u^2 = -1$$

Simplify the expression by subtracting \frac{225}{64} on both sides

$$-u^2 = -1-\frac{225}{64} = -\frac{289}{64}$$

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

$$u^2 = \frac{289}{64} u = \pm\sqrt{\frac{289}{64}} = \pm \frac{17}{8}$$

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

$$r =-\frac{15}{8} - \frac{17}{8} = -4 s = -\frac{15}{8} + \frac{17}{8} = 0.250$$

Evaluate

$$\left(4c-1\right)\left(c+4\right)$$