Solve 4x^2-5x-6

$$4 x ^ { 2 } - 5 x - 6$$

Factor

$$\left(x-2\right)\left(4x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=4\left(-6\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -5.

$$a=-8 b=3$$

Rewrite 4x^{2}-5x-6 as \left(4x^{2}-8x\right)+\left(3x-6\right).

$$\left(4x^{2}-8x\right)+\left(3x-6\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(x-2\right)+3\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(4x+3\right)$$

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.

$$4x^{2}-5x-6=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\left(-6\right)}}{2\times 4}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 4\left(-6\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-5\right)±\sqrt{25-16\left(-6\right)}}{2\times 4}$$

Multiply -16 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 4}$$

Add 25 to 96.

$$x=\frac{-\left(-5\right)±\sqrt{121}}{2\times 4}$$

Take the square root of 121.

$$x=\frac{-\left(-5\right)±11}{2\times 4}$$

The opposite of -5 is 5.

$$x=\frac{5±11}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{5±11}{8}$$

Now solve the equation x=\frac{5±11}{8} when ± is plus. Add 5 to 11.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{5±11}{8} when ± is minus. Subtract 11 from 5.

$$x=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -\frac{3}{4} for x_{2}.

$$4x^{2}-5x-6=4\left(x-2\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$4x^{2}-5x-6=4\left(x-2\right)\left(x+\frac{3}{4}\right)$$

Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-5x-6=4\left(x-2\right)\times \frac{4x+3}{4}$$

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

$$4x^{2}-5x-6=\left(x-2\right)\left(4x+3\right)$$

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 4

$$x ^ 2 -\frac{5}{4}x -\frac{3}{2} = 0$$

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

$$r + s = \frac{5}{4} rs = -\frac{3}{2}$$

Two numbers r and s sum up to \frac{5}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{4} = \frac{5}{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>

$$r = \frac{5}{8} - u s = \frac{5}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}

$$(\frac{5}{8} - u) (\frac{5}{8} + u) = -\frac{3}{2}$$

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

$$\frac{25}{64} - u^2 = -\frac{3}{2}$$

Simplify the expression by subtracting \frac{25}{64} on both sides

$$-u^2 = -\frac{3}{2}-\frac{25}{64} = -\frac{121}{64}$$

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

$$u^2 = \frac{121}{64} u = \pm\sqrt{\frac{121}{64}} = \pm \frac{11}{8}$$

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

$$r =\frac{5}{8} - \frac{11}{8} = -0.750 s = \frac{5}{8} + \frac{11}{8} = 2$$

Evaluate

$$\left(x-2\right)\left(4x+3\right)$$