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a+b=17 ab=3\times 10=30
Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+10. To find a and b, set up a system to be solved.
1,30 2,15 3,10 5,6
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 30.
1+30=31 2+15=17 3+10=13 5+6=11
Calculate the sum for each pair.
a=2 b=15
The solution is the pair that gives sum 17.
\left(3x^{2}+2x\right)+\left(15x+10\right)
Rewrite 3x^{2}+17x+10 as \left(3x^{2}+2x\right)+\left(15x+10\right).
x\left(3x+2\right)+5\left(3x+2\right)
Factor out x in the first and 5 in the second group.
\left(3x+2\right)\left(x+5\right)
Factor out common term 3x+2 by using distributive property.
3x^{2}+17x+10=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{-17±\sqrt{17^{2}-4\times 3\times 10}}{2\times 3}
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{-17±\sqrt{289-4\times 3\times 10}}{2\times 3}
Square 17.
x=\frac{-17±\sqrt{289-12\times 10}}{2\times 3}
Multiply -4 times 3.
x=\frac{-17±\sqrt{289-120}}{2\times 3}
Multiply -12 times 10.
x=\frac{-17±\sqrt{169}}{2\times 3}
Add 289 to -120.
x=\frac{-17±13}{2\times 3}
Take the square root of 169.
x=\frac{-17±13}{6}
Multiply 2 times 3.
x=-\frac{4}{6}
Now solve the equation x=\frac{-17±13}{6} when ± is plus. Add -17 to 13.
x=-\frac{2}{3}
Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.
x=-\frac{30}{6}
Now solve the equation x=\frac{-17±13}{6} when ± is minus. Subtract 13 from -17.
x=-5
Divide -30 by 6.
3x^{2}+17x+10=3\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-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{2}{3} for x_{1} and -5 for x_{2}.
3x^{2}+17x+10=3\left(x+\frac{2}{3}\right)\left(x+5\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
3x^{2}+17x+10=3\times \frac{3x+2}{3}\left(x+5\right)
Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
3x^{2}+17x+10=\left(3x+2\right)\left(x+5\right)
Cancel out 3, the greatest common factor in 3 and 3.
x ^ 2 +\frac{17}{3}x +\frac{10}{3} = 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 3
r + s = -\frac{17}{3} rs = \frac{10}{3}
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{17}{6} - u s = -\frac{17}{6} + u
Two numbers r and s sum up to -\frac{17}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{3} = -\frac{17}{6}. 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{17}{6} - u) (-\frac{17}{6} + u) = \frac{10}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{10}{3}
\frac{289}{36} - u^2 = \frac{10}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{10}{3}-\frac{289}{36} = -\frac{169}{36}
Simplify the expression by subtracting \frac{289}{36} on both sides
u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{17}{6} - \frac{13}{6} = -5 s = -\frac{17}{6} + \frac{13}{6} = -0.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.