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a+b=-43 ab=52\times 3=156
Factor the expression by grouping. First, the expression needs to be rewritten as 52z^{2}+az+bz+3. To find a and b, set up a system to be solved.
-1,-156 -2,-78 -3,-52 -4,-39 -6,-26 -12,-13
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 156.
-1-156=-157 -2-78=-80 -3-52=-55 -4-39=-43 -6-26=-32 -12-13=-25
Calculate the sum for each pair.
a=-39 b=-4
The solution is the pair that gives sum -43.
\left(52z^{2}-39z\right)+\left(-4z+3\right)
Rewrite 52z^{2}-43z+3 as \left(52z^{2}-39z\right)+\left(-4z+3\right).
13z\left(4z-3\right)-\left(4z-3\right)
Factor out 13z in the first and -1 in the second group.
\left(4z-3\right)\left(13z-1\right)
Factor out common term 4z-3 by using distributive property.
52z^{2}-43z+3=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.
z=\frac{-\left(-43\right)±\sqrt{\left(-43\right)^{2}-4\times 52\times 3}}{2\times 52}
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.
z=\frac{-\left(-43\right)±\sqrt{1849-4\times 52\times 3}}{2\times 52}
Square -43.
z=\frac{-\left(-43\right)±\sqrt{1849-208\times 3}}{2\times 52}
Multiply -4 times 52.
z=\frac{-\left(-43\right)±\sqrt{1849-624}}{2\times 52}
Multiply -208 times 3.
z=\frac{-\left(-43\right)±\sqrt{1225}}{2\times 52}
Add 1849 to -624.
z=\frac{-\left(-43\right)±35}{2\times 52}
Take the square root of 1225.
z=\frac{43±35}{2\times 52}
The opposite of -43 is 43.
z=\frac{43±35}{104}
Multiply 2 times 52.
z=\frac{78}{104}
Now solve the equation z=\frac{43±35}{104} when ± is plus. Add 43 to 35.
z=\frac{3}{4}
Reduce the fraction \frac{78}{104} to lowest terms by extracting and canceling out 26.
z=\frac{8}{104}
Now solve the equation z=\frac{43±35}{104} when ± is minus. Subtract 35 from 43.
z=\frac{1}{13}
Reduce the fraction \frac{8}{104} to lowest terms by extracting and canceling out 8.
52z^{2}-43z+3=52\left(z-\frac{3}{4}\right)\left(z-\frac{1}{13}\right)
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 \frac{1}{13} for x_{2}.
52z^{2}-43z+3=52\times \frac{4z-3}{4}\left(z-\frac{1}{13}\right)
Subtract \frac{3}{4} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
52z^{2}-43z+3=52\times \frac{4z-3}{4}\times \frac{13z-1}{13}
Subtract \frac{1}{13} from z by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
52z^{2}-43z+3=52\times \frac{\left(4z-3\right)\left(13z-1\right)}{4\times 13}
Multiply \frac{4z-3}{4} times \frac{13z-1}{13} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
52z^{2}-43z+3=52\times \frac{\left(4z-3\right)\left(13z-1\right)}{52}
Multiply 4 times 13.
52z^{2}-43z+3=\left(4z-3\right)\left(13z-1\right)
Cancel out 52, the greatest common factor in 52 and 52.
x ^ 2 -\frac{43}{52}x +\frac{3}{52} = 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 52
r + s = \frac{43}{52} rs = \frac{3}{52}
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{43}{104} - u s = \frac{43}{104} + u
Two numbers r and s sum up to \frac{43}{52} exactly when the average of the two numbers is \frac{1}{2}*\frac{43}{52} = \frac{43}{104}. 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{43}{104} - u) (\frac{43}{104} + u) = \frac{3}{52}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{52}
\frac{1849}{10816} - u^2 = \frac{3}{52}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{52}-\frac{1849}{10816} = -\frac{1225}{10816}
Simplify the expression by subtracting \frac{1849}{10816} on both sides
u^2 = \frac{1225}{10816} u = \pm\sqrt{\frac{1225}{10816}} = \pm \frac{35}{104}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{43}{104} - \frac{35}{104} = 0.07692307692307693 s = \frac{43}{104} + \frac{35}{104} = 0.75
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.