Factor
\left(5y-3\right)\left(2y+7\right)
Evaluate
\left(5y-3\right)\left(2y+7\right)
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a+b=29 ab=10\left(-21\right)=-210
Factor the expression by grouping. First, the expression needs to be rewritten as 10y^{2}+ay+by-21. To find a and b, set up a system to be solved.
-1,210 -2,105 -3,70 -5,42 -6,35 -7,30 -10,21 -14,15
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 -210.
-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1
Calculate the sum for each pair.
a=-6 b=35
The solution is the pair that gives sum 29.
\left(10y^{2}-6y\right)+\left(35y-21\right)
Rewrite 10y^{2}+29y-21 as \left(10y^{2}-6y\right)+\left(35y-21\right).
2y\left(5y-3\right)+7\left(5y-3\right)
Factor out 2y in the first and 7 in the second group.
\left(5y-3\right)\left(2y+7\right)
Factor out common term 5y-3 by using distributive property.
10y^{2}+29y-21=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.
y=\frac{-29±\sqrt{29^{2}-4\times 10\left(-21\right)}}{2\times 10}
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.
y=\frac{-29±\sqrt{841-4\times 10\left(-21\right)}}{2\times 10}
Square 29.
y=\frac{-29±\sqrt{841-40\left(-21\right)}}{2\times 10}
Multiply -4 times 10.
y=\frac{-29±\sqrt{841+840}}{2\times 10}
Multiply -40 times -21.
y=\frac{-29±\sqrt{1681}}{2\times 10}
Add 841 to 840.
y=\frac{-29±41}{2\times 10}
Take the square root of 1681.
y=\frac{-29±41}{20}
Multiply 2 times 10.
y=\frac{12}{20}
Now solve the equation y=\frac{-29±41}{20} when ± is plus. Add -29 to 41.
y=\frac{3}{5}
Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.
y=-\frac{70}{20}
Now solve the equation y=\frac{-29±41}{20} when ± is minus. Subtract 41 from -29.
y=-\frac{7}{2}
Reduce the fraction \frac{-70}{20} to lowest terms by extracting and canceling out 10.
10y^{2}+29y-21=10\left(y-\frac{3}{5}\right)\left(y-\left(-\frac{7}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{5} for x_{1} and -\frac{7}{2} for x_{2}.
10y^{2}+29y-21=10\left(y-\frac{3}{5}\right)\left(y+\frac{7}{2}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
10y^{2}+29y-21=10\times \frac{5y-3}{5}\left(y+\frac{7}{2}\right)
Subtract \frac{3}{5} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
10y^{2}+29y-21=10\times \frac{5y-3}{5}\times \frac{2y+7}{2}
Add \frac{7}{2} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
10y^{2}+29y-21=10\times \frac{\left(5y-3\right)\left(2y+7\right)}{5\times 2}
Multiply \frac{5y-3}{5} times \frac{2y+7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
10y^{2}+29y-21=10\times \frac{\left(5y-3\right)\left(2y+7\right)}{10}
Multiply 5 times 2.
10y^{2}+29y-21=\left(5y-3\right)\left(2y+7\right)
Cancel out 10, the greatest common factor in 10 and 10.
x ^ 2 +\frac{29}{10}x -\frac{21}{10} = 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 10
r + s = -\frac{29}{10} rs = -\frac{21}{10}
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{29}{20} - u s = -\frac{29}{20} + u
Two numbers r and s sum up to -\frac{29}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{29}{10} = -\frac{29}{20}. 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{29}{20} - u) (-\frac{29}{20} + u) = -\frac{21}{10}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{21}{10}
\frac{841}{400} - u^2 = -\frac{21}{10}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{21}{10}-\frac{841}{400} = -\frac{1681}{400}
Simplify the expression by subtracting \frac{841}{400} on both sides
u^2 = \frac{1681}{400} u = \pm\sqrt{\frac{1681}{400}} = \pm \frac{41}{20}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{29}{20} - \frac{41}{20} = -3.500 s = -\frac{29}{20} + \frac{41}{20} = 0.600
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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