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a+b=7 ab=5\times 2=10
Factor the expression by grouping. First, the expression needs to be rewritten as 5z^{2}+az+bz+2. To find a and b, set up a system to be solved.
1,10 2,5
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 10.
1+10=11 2+5=7
Calculate the sum for each pair.
a=2 b=5
The solution is the pair that gives sum 7.
\left(5z^{2}+2z\right)+\left(5z+2\right)
Rewrite 5z^{2}+7z+2 as \left(5z^{2}+2z\right)+\left(5z+2\right).
z\left(5z+2\right)+5z+2
Factor out z in 5z^{2}+2z.
\left(5z+2\right)\left(z+1\right)
Factor out common term 5z+2 by using distributive property.
5z^{2}+7z+2=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{-7±\sqrt{7^{2}-4\times 5\times 2}}{2\times 5}
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{-7±\sqrt{49-4\times 5\times 2}}{2\times 5}
Square 7.
z=\frac{-7±\sqrt{49-20\times 2}}{2\times 5}
Multiply -4 times 5.
z=\frac{-7±\sqrt{49-40}}{2\times 5}
Multiply -20 times 2.
z=\frac{-7±\sqrt{9}}{2\times 5}
Add 49 to -40.
z=\frac{-7±3}{2\times 5}
Take the square root of 9.
z=\frac{-7±3}{10}
Multiply 2 times 5.
z=-\frac{4}{10}
Now solve the equation z=\frac{-7±3}{10} when ± is plus. Add -7 to 3.
z=-\frac{2}{5}
Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.
z=-\frac{10}{10}
Now solve the equation z=\frac{-7±3}{10} when ± is minus. Subtract 3 from -7.
z=-1
Divide -10 by 10.
5z^{2}+7z+2=5\left(z-\left(-\frac{2}{5}\right)\right)\left(z-\left(-1\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{5} for x_{1} and -1 for x_{2}.
5z^{2}+7z+2=5\left(z+\frac{2}{5}\right)\left(z+1\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
5z^{2}+7z+2=5\times \frac{5z+2}{5}\left(z+1\right)
Add \frac{2}{5} to z by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
5z^{2}+7z+2=\left(5z+2\right)\left(z+1\right)
Cancel out 5, the greatest common factor in 5 and 5.
x ^ 2 +\frac{7}{5}x +\frac{2}{5} = 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 5
r + s = -\frac{7}{5} rs = \frac{2}{5}
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{7}{10} - u s = -\frac{7}{10} + u
Two numbers r and s sum up to -\frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{5} = -\frac{7}{10}. 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{7}{10} - u) (-\frac{7}{10} + u) = \frac{2}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{5}
\frac{49}{100} - u^2 = \frac{2}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{5}-\frac{49}{100} = -\frac{9}{100}
Simplify the expression by subtracting \frac{49}{100} on both sides
u^2 = \frac{9}{100} u = \pm\sqrt{\frac{9}{100}} = \pm \frac{3}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{7}{10} - \frac{3}{10} = -1.000 s = -\frac{7}{10} + \frac{3}{10} = -0.400
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.