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a+b=43 ab=35\left(-36\right)=-1260
Factor the expression by grouping. First, the expression needs to be rewritten as 35x^{2}+ax+bx-36. To find a and b, set up a system to be solved.
-1,1260 -2,630 -3,420 -4,315 -5,252 -6,210 -7,180 -9,140 -10,126 -12,105 -14,90 -15,84 -18,70 -20,63 -21,60 -28,45 -30,42 -35,36
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -1260.
-1+1260=1259 -2+630=628 -3+420=417 -4+315=311 -5+252=247 -6+210=204 -7+180=173 -9+140=131 -10+126=116 -12+105=93 -14+90=76 -15+84=69 -18+70=52 -20+63=43 -21+60=39 -28+45=17 -30+42=12 -35+36=1
Calculate the sum for each pair.
a=-20 b=63
The solution is the pair that gives sum 43.
\left(35x^{2}-20x\right)+\left(63x-36\right)
Rewrite 35x^{2}+43x-36 as \left(35x^{2}-20x\right)+\left(63x-36\right).
5x\left(7x-4\right)+9\left(7x-4\right)
Factor out 5x in the first and 9 in the second group.
\left(7x-4\right)\left(5x+9\right)
Factor out common term 7x-4 by using distributive property.
35x^{2}+43x-36=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{-43±\sqrt{43^{2}-4\times 35\left(-36\right)}}{2\times 35}
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{-43±\sqrt{1849-4\times 35\left(-36\right)}}{2\times 35}
Square 43.
x=\frac{-43±\sqrt{1849-140\left(-36\right)}}{2\times 35}
Multiply -4 times 35.
x=\frac{-43±\sqrt{1849+5040}}{2\times 35}
Multiply -140 times -36.
x=\frac{-43±\sqrt{6889}}{2\times 35}
Add 1849 to 5040.
x=\frac{-43±83}{2\times 35}
Take the square root of 6889.
x=\frac{-43±83}{70}
Multiply 2 times 35.
x=\frac{40}{70}
Now solve the equation x=\frac{-43±83}{70} when ± is plus. Add -43 to 83.
x=\frac{4}{7}
Reduce the fraction \frac{40}{70} to lowest terms by extracting and canceling out 10.
x=-\frac{126}{70}
Now solve the equation x=\frac{-43±83}{70} when ± is minus. Subtract 83 from -43.
x=-\frac{9}{5}
Reduce the fraction \frac{-126}{70} to lowest terms by extracting and canceling out 14.
35x^{2}+43x-36=35\left(x-\frac{4}{7}\right)\left(x-\left(-\frac{9}{5}\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}{7} for x_{1} and -\frac{9}{5} for x_{2}.
35x^{2}+43x-36=35\left(x-\frac{4}{7}\right)\left(x+\frac{9}{5}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
35x^{2}+43x-36=35\times \frac{7x-4}{7}\left(x+\frac{9}{5}\right)
Subtract \frac{4}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
35x^{2}+43x-36=35\times \frac{7x-4}{7}\times \frac{5x+9}{5}
Add \frac{9}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
35x^{2}+43x-36=35\times \frac{\left(7x-4\right)\left(5x+9\right)}{7\times 5}
Multiply \frac{7x-4}{7} times \frac{5x+9}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
35x^{2}+43x-36=35\times \frac{\left(7x-4\right)\left(5x+9\right)}{35}
Multiply 7 times 5.
35x^{2}+43x-36=\left(7x-4\right)\left(5x+9\right)
Cancel out 35, the greatest common factor in 35 and 35.
x ^ 2 +\frac{43}{35}x -\frac{36}{35} = 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 35
r + s = -\frac{43}{35} rs = -\frac{36}{35}
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}{70} - u s = -\frac{43}{70} + u
Two numbers r and s sum up to -\frac{43}{35} exactly when the average of the two numbers is \frac{1}{2}*-\frac{43}{35} = -\frac{43}{70}. 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}{70} - u) (-\frac{43}{70} + u) = -\frac{36}{35}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{36}{35}
\frac{1849}{4900} - u^2 = -\frac{36}{35}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{36}{35}-\frac{1849}{4900} = -\frac{6889}{4900}
Simplify the expression by subtracting \frac{1849}{4900} on both sides
u^2 = \frac{6889}{4900} u = \pm\sqrt{\frac{6889}{4900}} = \pm \frac{83}{70}
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}{70} - \frac{83}{70} = -1.800 s = -\frac{43}{70} + \frac{83}{70} = 0.571
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.