Factor
\left(a-3\right)\left(5a-1\right)
Evaluate
\left(a-3\right)\left(5a-1\right)
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p+q=-16 pq=5\times 3=15
Factor the expression by grouping. First, the expression needs to be rewritten as 5a^{2}+pa+qa+3. To find p and q, set up a system to be solved.
-1,-15 -3,-5
Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 15.
-1-15=-16 -3-5=-8
Calculate the sum for each pair.
p=-15 q=-1
The solution is the pair that gives sum -16.
\left(5a^{2}-15a\right)+\left(-a+3\right)
Rewrite 5a^{2}-16a+3 as \left(5a^{2}-15a\right)+\left(-a+3\right).
5a\left(a-3\right)-\left(a-3\right)
Factor out 5a in the first and -1 in the second group.
\left(a-3\right)\left(5a-1\right)
Factor out common term a-3 by using distributive property.
5a^{2}-16a+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.
a=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 3}}{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.
a=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 3}}{2\times 5}
Square -16.
a=\frac{-\left(-16\right)±\sqrt{256-20\times 3}}{2\times 5}
Multiply -4 times 5.
a=\frac{-\left(-16\right)±\sqrt{256-60}}{2\times 5}
Multiply -20 times 3.
a=\frac{-\left(-16\right)±\sqrt{196}}{2\times 5}
Add 256 to -60.
a=\frac{-\left(-16\right)±14}{2\times 5}
Take the square root of 196.
a=\frac{16±14}{2\times 5}
The opposite of -16 is 16.
a=\frac{16±14}{10}
Multiply 2 times 5.
a=\frac{30}{10}
Now solve the equation a=\frac{16±14}{10} when ± is plus. Add 16 to 14.
a=3
Divide 30 by 10.
a=\frac{2}{10}
Now solve the equation a=\frac{16±14}{10} when ± is minus. Subtract 14 from 16.
a=\frac{1}{5}
Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.
5a^{2}-16a+3=5\left(a-3\right)\left(a-\frac{1}{5}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{1}{5} for x_{2}.
5a^{2}-16a+3=5\left(a-3\right)\times \frac{5a-1}{5}
Subtract \frac{1}{5} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
5a^{2}-16a+3=\left(a-3\right)\left(5a-1\right)
Cancel out 5, the greatest common factor in 5 and 5.
x ^ 2 -\frac{16}{5}x +\frac{3}{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{16}{5} rs = \frac{3}{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{8}{5} - u s = \frac{8}{5} + u
Two numbers r and s sum up to \frac{16}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{5} = \frac{8}{5}. 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{8}{5} - u) (\frac{8}{5} + u) = \frac{3}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{5}
\frac{64}{25} - u^2 = \frac{3}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{5}-\frac{64}{25} = -\frac{49}{25}
Simplify the expression by subtracting \frac{64}{25} on both sides
u^2 = \frac{49}{25} u = \pm\sqrt{\frac{49}{25}} = \pm \frac{7}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{8}{5} - \frac{7}{5} = 0.200 s = \frac{8}{5} + \frac{7}{5} = 3
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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