Solve 24t^2+19t+2=

$$24 t ^ { 2 } + 19 t + 2 =$$

Factor

$$\left(3t+2\right)\left(8t+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24t^{2}+at+bt+2. To find a and b, set up a system to be solved.

$$a+b=19 ab=24\times 2=48$$

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 48.

$$1,48 2,24 3,16 4,12 6,8$$

Calculate the sum for each pair.

$$1+48=49 2+24=26 3+16=19 4+12=16 6+8=14$$

The solution is the pair that gives sum 19.

$$a=3 b=16$$

Rewrite 24t^{2}+19t+2 as \left(24t^{2}+3t\right)+\left(16t+2\right).

$$\left(24t^{2}+3t\right)+\left(16t+2\right)$$

Factor out 3t in the first and 2 in the second group.

$$3t\left(8t+1\right)+2\left(8t+1\right)$$

Factor out common term 8t+1 by using distributive property.

$$\left(8t+1\right)\left(3t+2\right)$$

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.

$$24t^{2}+19t+2=0$$

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.

$$t=\frac{-19±\sqrt{19^{2}-4\times 24\times 2}}{2\times 24}$$

Square 19.

$$t=\frac{-19±\sqrt{361-4\times 24\times 2}}{2\times 24}$$

Multiply -4 times 24.

$$t=\frac{-19±\sqrt{361-96\times 2}}{2\times 24}$$

Multiply -96 times 2.

$$t=\frac{-19±\sqrt{361-192}}{2\times 24}$$

Add 361 to -192.

$$t=\frac{-19±\sqrt{169}}{2\times 24}$$

Take the square root of 169.

$$t=\frac{-19±13}{2\times 24}$$

Multiply 2 times 24.

$$t=\frac{-19±13}{48}$$

Now solve the equation t=\frac{-19±13}{48} when ± is plus. Add -19 to 13.

$$t=-\frac{6}{48}$$

Reduce the fraction \frac{-6}{48} to lowest terms by extracting and canceling out 6.

$$t=-\frac{1}{8}$$

Now solve the equation t=\frac{-19±13}{48} when ± is minus. Subtract 13 from -19.

$$t=-\frac{32}{48}$$

Reduce the fraction \frac{-32}{48} to lowest terms by extracting and canceling out 16.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{8} for x_{1} and -\frac{2}{3} for x_{2}.

$$24t^{2}+19t+2=24\left(t-\left(-\frac{1}{8}\right)\right)\left(t-\left(-\frac{2}{3}\right)\right)$$

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

$$24t^{2}+19t+2=24\left(t+\frac{1}{8}\right)\left(t+\frac{2}{3}\right)$$

Add \frac{1}{8} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$24t^{2}+19t+2=24\times \frac{8t+1}{8}\left(t+\frac{2}{3}\right)$$

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

$$24t^{2}+19t+2=24\times \frac{8t+1}{8}\times \frac{3t+2}{3}$$

Multiply \frac{8t+1}{8} times \frac{3t+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24t^{2}+19t+2=24\times \frac{\left(8t+1\right)\left(3t+2\right)}{8\times 3}$$

Multiply 8 times 3.

$$24t^{2}+19t+2=24\times \frac{\left(8t+1\right)\left(3t+2\right)}{24}$$

Cancel out 24, the greatest common factor in 24 and 24.

$$24t^{2}+19t+2=\left(8t+1\right)\left(3t+2\right)$$

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 24

$$x ^ 2 +\frac{19}{24}x +\frac{1}{12} = 0$$

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

$$r + s = -\frac{19}{24} rs = \frac{1}{12}$$

Two numbers r and s sum up to -\frac{19}{24} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{24} = -\frac{19}{48}. 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>

$$r = -\frac{19}{48} - u s = -\frac{19}{48} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{12}

$$(-\frac{19}{48} - u) (-\frac{19}{48} + u) = \frac{1}{12}$$

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

$$\frac{361}{2304} - u^2 = \frac{1}{12}$$

Simplify the expression by subtracting \frac{361}{2304} on both sides

$$-u^2 = \frac{1}{12}-\frac{361}{2304} = -\frac{169}{2304}$$

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

$$u^2 = \frac{169}{2304} u = \pm\sqrt{\frac{169}{2304}} = \pm \frac{13}{48}$$

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

$$r =-\frac{19}{48} - \frac{13}{48} = -0.667 s = -\frac{19}{48} + \frac{13}{48} = -0.125$$

Evaluate

$$\left(3t+2\right)\left(8t+1\right)$$