Solve 12x^2+69x+45

$$12 x ^ { 2 } + 69 x + 45$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(4x^{2}+23x+15\right)$$

Consider 4x^{2}+23x+15. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=23 ab=4\times 15=60$$

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 4x^{2}+23x+15 as \left(4x^{2}+3x\right)+\left(20x+15\right).

$$\left(4x^{2}+3x\right)+\left(20x+15\right)$$

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

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

Factor out common term 4x+3 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$12x^{2}+69x+45=0$$

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.

$$x=\frac{-69±\sqrt{69^{2}-4\times 12\times 45}}{2\times 12}$$

Square 69.

$$x=\frac{-69±\sqrt{4761-4\times 12\times 45}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-69±\sqrt{4761-48\times 45}}{2\times 12}$$

Multiply -48 times 45.

$$x=\frac{-69±\sqrt{4761-2160}}{2\times 12}$$

Add 4761 to -2160.

$$x=\frac{-69±\sqrt{2601}}{2\times 12}$$

Take the square root of 2601.

$$x=\frac{-69±51}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-69±51}{24}$$

Now solve the equation x=\frac{-69±51}{24} when ± is plus. Add -69 to 51.

$$x=-\frac{18}{24}$$

Reduce the fraction \frac{-18}{24} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-69±51}{24} when ± is minus. Subtract 51 from -69.

$$x=-\frac{120}{24}$$

Divide -120 by 24.

$$x=-5$$

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

$$12x^{2}+69x+45=12\left(x-\left(-\frac{3}{4}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$12x^{2}+69x+45=12\left(x+\frac{3}{4}\right)\left(x+5\right)$$

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

$$12x^{2}+69x+45=12\times \frac{4x+3}{4}\left(x+5\right)$$

Cancel out 4, the greatest common factor in 12 and 4.

$$12x^{2}+69x+45=3\left(4x+3\right)\left(x+5\right)$$

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

$$x ^ 2 +\frac{23}{4}x +\frac{15}{4} = 0$$

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

$$r + s = -\frac{23}{4} rs = \frac{15}{4}$$

Two numbers r and s sum up to -\frac{23}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{23}{4} = -\frac{23}{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>

$$r = -\frac{23}{8} - u s = -\frac{23}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{4}

$$(-\frac{23}{8} - u) (-\frac{23}{8} + u) = \frac{15}{4}$$

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

$$\frac{529}{64} - u^2 = \frac{15}{4}$$

Simplify the expression by subtracting \frac{529}{64} on both sides

$$-u^2 = \frac{15}{4}-\frac{529}{64} = -\frac{289}{64}$$

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.

$$r =-\frac{23}{8} - \frac{17}{8} = -5 s = -\frac{23}{8} + \frac{17}{8} = -0.750$$

Evaluate

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