Solve 12x^2+23x+5

$$12 x ^ { 2 } + 23 x + 5$$

Factor

$$\left(3x+5\right)\left(4x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=23 ab=12\times 5=60$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 23.

$$a=3 b=20$$

Rewrite 12x^{2}+23x+5 as \left(12x^{2}+3x\right)+\left(20x+5\right).

$$\left(12x^{2}+3x\right)+\left(20x+5\right)$$

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

$$3x\left(4x+1\right)+5\left(4x+1\right)$$

Factor out common term 4x+1 by using distributive property.

$$\left(4x+1\right)\left(3x+5\right)$$

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.

$$12x^{2}+23x+5=0$$

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.

$$x=\frac{-23±\sqrt{23^{2}-4\times 12\times 5}}{2\times 12}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\times 12\times 5}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-23±\sqrt{529-48\times 5}}{2\times 12}$$

Multiply -48 times 5.

$$x=\frac{-23±\sqrt{529-240}}{2\times 12}$$

Add 529 to -240.

$$x=\frac{-23±\sqrt{289}}{2\times 12}$$

Take the square root of 289.

$$x=\frac{-23±17}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-23±17}{24}$$

Now solve the equation x=\frac{-23±17}{24} when ± is plus. Add -23 to 17.

$$x=-\frac{6}{24}$$

Reduce the fraction \frac{-6}{24} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{4}$$

Now solve the equation x=\frac{-23±17}{24} when ± is minus. Subtract 17 from -23.

$$x=-\frac{40}{24}$$

Reduce the fraction \frac{-40}{24} to lowest terms by extracting and canceling out 8.

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

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 -\frac{5}{3} for x_{2}.

$$12x^{2}+23x+5=12\left(x-\left(-\frac{1}{4}\right)\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$12x^{2}+23x+5=12\left(x+\frac{1}{4}\right)\left(x+\frac{5}{3}\right)$$

Add \frac{1}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+23x+5=12\times \frac{4x+1}{4}\left(x+\frac{5}{3}\right)$$

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

$$12x^{2}+23x+5=12\times \frac{4x+1}{4}\times \frac{3x+5}{3}$$

Multiply \frac{4x+1}{4} times \frac{3x+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+23x+5=12\times \frac{\left(4x+1\right)\left(3x+5\right)}{4\times 3}$$

Multiply 4 times 3.

$$12x^{2}+23x+5=12\times \frac{\left(4x+1\right)\left(3x+5\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12x^{2}+23x+5=\left(4x+1\right)\left(3x+5\right)$$

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 12

$$x ^ 2 +\frac{23}{12}x +\frac{5}{12} = 0$$

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

$$r + s = -\frac{23}{12} rs = \frac{5}{12}$$

Two numbers r and s sum up to -\frac{23}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{23}{12} = -\frac{23}{24}. 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>

$$r = -\frac{23}{24} - u s = -\frac{23}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{12}

$$(-\frac{23}{24} - u) (-\frac{23}{24} + u) = \frac{5}{12}$$

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

$$\frac{529}{576} - u^2 = \frac{5}{12}$$

Simplify the expression by subtracting \frac{529}{576} on both sides

$$-u^2 = \frac{5}{12}-\frac{529}{576} = -\frac{289}{576}$$

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

$$u^2 = \frac{289}{576} u = \pm\sqrt{\frac{289}{576}} = \pm \frac{17}{24}$$

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

$$r =-\frac{23}{24} - \frac{17}{24} = -1.667 s = -\frac{23}{24} + \frac{17}{24} = -0.250$$

Evaluate

$$\left(3x+5\right)\left(4x+1\right)$$