Solve 12x^2+23x+10

$$12 x ^ { 2 } + 23 x + 10$$

Factor

$$\left(3x+2\right)\left(4x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=23 ab=12\times 10=120$$

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 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 23.

$$a=8 b=15$$

Rewrite 12x^{2}+23x+10 as \left(12x^{2}+8x\right)+\left(15x+10\right).

$$\left(12x^{2}+8x\right)+\left(15x+10\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(3x+2\right)+5\left(3x+2\right)$$

Factor out common term 3x+2 by using distributive property.

$$\left(3x+2\right)\left(4x+5\right)$$

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.

$$12x^{2}+23x+10=0$$

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.

$$x=\frac{-23±\sqrt{23^{2}-4\times 12\times 10}}{2\times 12}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\times 12\times 10}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-23±\sqrt{529-48\times 10}}{2\times 12}$$

Multiply -48 times 10.

$$x=\frac{-23±\sqrt{529-480}}{2\times 12}$$

Add 529 to -480.

$$x=\frac{-23±\sqrt{49}}{2\times 12}$$

Take the square root of 49.

$$x=\frac{-23±7}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-23±7}{24}$$

Now solve the equation x=\frac{-23±7}{24} when ± is plus. Add -23 to 7.

$$x=-\frac{16}{24}$$

Reduce the fraction \frac{-16}{24} to lowest terms by extracting and canceling out 8.

$$x=-\frac{2}{3}$$

Now solve the equation x=\frac{-23±7}{24} when ± is minus. Subtract 7 from -23.

$$x=-\frac{30}{24}$$

Reduce the fraction \frac{-30}{24} to lowest terms by extracting and canceling out 6.

$$x=-\frac{5}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3} for x_{1} and -\frac{5}{4} for x_{2}.

$$12x^{2}+23x+10=12\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-\frac{5}{4}\right)\right)$$

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

$$12x^{2}+23x+10=12\left(x+\frac{2}{3}\right)\left(x+\frac{5}{4}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+23x+10=12\times \frac{3x+2}{3}\left(x+\frac{5}{4}\right)$$

Add \frac{5}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+23x+10=12\times \frac{3x+2}{3}\times \frac{4x+5}{4}$$

Multiply \frac{3x+2}{3} times \frac{4x+5}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+23x+10=12\times \frac{\left(3x+2\right)\left(4x+5\right)}{3\times 4}$$

Multiply 3 times 4.

$$12x^{2}+23x+10=12\times \frac{\left(3x+2\right)\left(4x+5\right)}{12}$$

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

$$12x^{2}+23x+10=\left(3x+2\right)\left(4x+5\right)$$

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

$$x ^ 2 +\frac{23}{12}x +\frac{5}{6} = 0$$

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

$$r + s = -\frac{23}{12} rs = \frac{5}{6}$$

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}{6}

$$(-\frac{23}{24} - u) (-\frac{23}{24} + u) = \frac{5}{6}$$

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

$$\frac{529}{576} - u^2 = \frac{5}{6}$$

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

$$-u^2 = \frac{5}{6}-\frac{529}{576} = -\frac{49}{576}$$

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

$$u^2 = \frac{49}{576} u = \pm\sqrt{\frac{49}{576}} = \pm \frac{7}{24}$$

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

$$r =-\frac{23}{24} - \frac{7}{24} = -1.250 s = -\frac{23}{24} + \frac{7}{24} = -0.667$$

Evaluate

$$\left(3x+2\right)\left(4x+5\right)$$