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3\left(x^{2}-11x+30\right)
Factor out 3.
a+b=-11 ab=1\times 30=30
Consider x^{2}-11x+30. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+30. 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 negative, a and b are both negative. 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=-6 b=-5
The solution is the pair that gives sum -11.
\left(x^{2}-6x\right)+\left(-5x+30\right)
Rewrite x^{2}-11x+30 as \left(x^{2}-6x\right)+\left(-5x+30\right).
x\left(x-6\right)-5\left(x-6\right)
Factor out x in the first and -5 in the second group.
\left(x-6\right)\left(x-5\right)
Factor out common term x-6 by using distributive property.
3\left(x-6\right)\left(x-5\right)
Rewrite the complete factored expression.
3x^{2}-33x+90=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{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 3\times 90}}{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{-\left(-33\right)±\sqrt{1089-4\times 3\times 90}}{2\times 3}
Square -33.
x=\frac{-\left(-33\right)±\sqrt{1089-12\times 90}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-33\right)±\sqrt{1089-1080}}{2\times 3}
Multiply -12 times 90.
x=\frac{-\left(-33\right)±\sqrt{9}}{2\times 3}
Add 1089 to -1080.
x=\frac{-\left(-33\right)±3}{2\times 3}
Take the square root of 9.
x=\frac{33±3}{2\times 3}
The opposite of -33 is 33.
x=\frac{33±3}{6}
Multiply 2 times 3.
x=\frac{36}{6}
Now solve the equation x=\frac{33±3}{6} when ± is plus. Add 33 to 3.
x=6
Divide 36 by 6.
x=\frac{30}{6}
Now solve the equation x=\frac{33±3}{6} when ± is minus. Subtract 3 from 33.
x=5
Divide 30 by 6.
3x^{2}-33x+90=3\left(x-6\right)\left(x-5\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and 5 for x_{2}.
x ^ 2 -11x +30 = 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 = 11 rs = 30
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{11}{2} - u s = \frac{11}{2} + u
Two numbers r and s sum up to 11 exactly when the average of the two numbers is \frac{1}{2}*11 = \frac{11}{2}. 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{11}{2} - u) (\frac{11}{2} + u) = 30
To solve for unknown quantity u, substitute these in the product equation rs = 30
\frac{121}{4} - u^2 = 30
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 30-\frac{121}{4} = -\frac{1}{4}
Simplify the expression by subtracting \frac{121}{4} on both sides
u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{11}{2} - \frac{1}{2} = 5 s = \frac{11}{2} + \frac{1}{2} = 6
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.