Factor
\left(3x+4\right)\left(4x+7\right)
Evaluate
\left(3x+4\right)\left(4x+7\right)
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a+b=37 ab=12\times 28=336
Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+28. To find a and b, set up a system to be solved.
1,336 2,168 3,112 4,84 6,56 7,48 8,42 12,28 14,24 16,21
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 336.
1+336=337 2+168=170 3+112=115 4+84=88 6+56=62 7+48=55 8+42=50 12+28=40 14+24=38 16+21=37
Calculate the sum for each pair.
a=16 b=21
The solution is the pair that gives sum 37.
\left(12x^{2}+16x\right)+\left(21x+28\right)
Rewrite 12x^{2}+37x+28 as \left(12x^{2}+16x\right)+\left(21x+28\right).
4x\left(3x+4\right)+7\left(3x+4\right)
Factor out 4x in the first and 7 in the second group.
\left(3x+4\right)\left(4x+7\right)
Factor out common term 3x+4 by using distributive property.
12x^{2}+37x+28=0
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.
x=\frac{-37±\sqrt{37^{2}-4\times 12\times 28}}{2\times 12}
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{-37±\sqrt{1369-4\times 12\times 28}}{2\times 12}
Square 37.
x=\frac{-37±\sqrt{1369-48\times 28}}{2\times 12}
Multiply -4 times 12.
x=\frac{-37±\sqrt{1369-1344}}{2\times 12}
Multiply -48 times 28.
x=\frac{-37±\sqrt{25}}{2\times 12}
Add 1369 to -1344.
x=\frac{-37±5}{2\times 12}
Take the square root of 25.
x=\frac{-37±5}{24}
Multiply 2 times 12.
x=-\frac{32}{24}
Now solve the equation x=\frac{-37±5}{24} when ± is plus. Add -37 to 5.
x=-\frac{4}{3}
Reduce the fraction \frac{-32}{24} to lowest terms by extracting and canceling out 8.
x=-\frac{42}{24}
Now solve the equation x=\frac{-37±5}{24} when ± is minus. Subtract 5 from -37.
x=-\frac{7}{4}
Reduce the fraction \frac{-42}{24} to lowest terms by extracting and canceling out 6.
12x^{2}+37x+28=12\left(x-\left(-\frac{4}{3}\right)\right)\left(x-\left(-\frac{7}{4}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{4}{3} for x_{1} and -\frac{7}{4} for x_{2}.
12x^{2}+37x+28=12\left(x+\frac{4}{3}\right)\left(x+\frac{7}{4}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
12x^{2}+37x+28=12\times \frac{3x+4}{3}\left(x+\frac{7}{4}\right)
Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
12x^{2}+37x+28=12\times \frac{3x+4}{3}\times \frac{4x+7}{4}
Add \frac{7}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
12x^{2}+37x+28=12\times \frac{\left(3x+4\right)\left(4x+7\right)}{3\times 4}
Multiply \frac{3x+4}{3} times \frac{4x+7}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
12x^{2}+37x+28=12\times \frac{\left(3x+4\right)\left(4x+7\right)}{12}
Multiply 3 times 4.
12x^{2}+37x+28=\left(3x+4\right)\left(4x+7\right)
Cancel out 12, the greatest common factor in 12 and 12.
x ^ 2 +\frac{37}{12}x +\frac{7}{3} = 0
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
r + s = -\frac{37}{12} rs = \frac{7}{3}
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 = -\frac{37}{24} - u s = -\frac{37}{24} + u
Two numbers r and s sum up to -\frac{37}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{37}{12} = -\frac{37}{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>
(-\frac{37}{24} - u) (-\frac{37}{24} + u) = \frac{7}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{3}
\frac{1369}{576} - u^2 = \frac{7}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{7}{3}-\frac{1369}{576} = -\frac{25}{576}
Simplify the expression by subtracting \frac{1369}{576} on both sides
u^2 = \frac{25}{576} u = \pm\sqrt{\frac{25}{576}} = \pm \frac{5}{24}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{37}{24} - \frac{5}{24} = -1.750 s = -\frac{37}{24} + \frac{5}{24} = -1.333
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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