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3x^{2}+18x+1=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{-18±\sqrt{18^{2}-4\times 3}}{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{-18±\sqrt{324-4\times 3}}{2\times 3}
Square 18.
x=\frac{-18±\sqrt{324-12}}{2\times 3}
Multiply -4 times 3.
x=\frac{-18±\sqrt{312}}{2\times 3}
Add 324 to -12.
x=\frac{-18±2\sqrt{78}}{2\times 3}
Take the square root of 312.
x=\frac{-18±2\sqrt{78}}{6}
Multiply 2 times 3.
x=\frac{2\sqrt{78}-18}{6}
Now solve the equation x=\frac{-18±2\sqrt{78}}{6} when ± is plus. Add -18 to 2\sqrt{78}.
x=\frac{\sqrt{78}}{3}-3
Divide -18+2\sqrt{78} by 6.
x=\frac{-2\sqrt{78}-18}{6}
Now solve the equation x=\frac{-18±2\sqrt{78}}{6} when ± is minus. Subtract 2\sqrt{78} from -18.
x=-\frac{\sqrt{78}}{3}-3
Divide -18-2\sqrt{78} by 6.
3x^{2}+18x+1=3\left(x-\left(\frac{\sqrt{78}}{3}-3\right)\right)\left(x-\left(-\frac{\sqrt{78}}{3}-3\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\frac{\sqrt{78}}{3} for x_{1} and -3-\frac{\sqrt{78}}{3} for x_{2}.