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Solution Steps
Use binomial theorem to expand .
To find the opposite of , find the opposite of each term.
Subtract from to get .
Expand
Solution Steps
Use binomial theorem to expand .
To find the opposite of , find the opposite of each term.
Subtract from to get .
Factor
Steps Using the Quadratic Formula
Use binomial theorem to expand .
To find the opposite of , find the opposite of each term.
Subtract from to get .
Quadratic polynomial can be factored using the transformation , where and are the solutions of the quadratic equation .
All equations of the form can be solved using the quadratic formula: . The quadratic formula gives two solutions, one when is addition and one when it is subtraction.
Square .
Multiply times .
Multiply times .
Add to .
Take the square root of .
Multiply times .
Now solve the equation when is plus. Add to .
Divide by .
Now solve the equation when is minus. Subtract from .
Divide by .
Factor the original expression using . Substitute for and for .
Graph
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Similar Problems from Web Search

5-\left(x^{2}-8x+16\right)
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.
5-x^{2}+8x-16
To find the opposite of x^{2}-8x+16, find the opposite of each term.
-11-x^{2}+8x
Subtract 16 from 5 to get -11.
5-\left(x^{2}-8x+16\right)
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.
5-x^{2}+8x-16
To find the opposite of x^{2}-8x+16, find the opposite of each term.
-11-x^{2}+8x
Subtract 16 from 5 to get -11.
factor(5-\left(x^{2}-8x+16\right))
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.
factor(5-x^{2}+8x-16)
To find the opposite of x^{2}-8x+16, find the opposite of each term.
factor(-11-x^{2}+8x)
Subtract 16 from 5 to get -11.
-x^{2}+8x-11=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{-8±\sqrt{8^{2}-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}
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{-8±\sqrt{64-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}
Square 8.
x=\frac{-8±\sqrt{64+4\left(-11\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-8±\sqrt{64-44}}{2\left(-1\right)}
Multiply 4 times -11.
x=\frac{-8±\sqrt{20}}{2\left(-1\right)}
Add 64 to -44.
x=\frac{-8±2\sqrt{5}}{2\left(-1\right)}
Take the square root of 20.
x=\frac{-8±2\sqrt{5}}{-2}
Multiply 2 times -1.
x=\frac{2\sqrt{5}-8}{-2}
Now solve the equation x=\frac{-8±2\sqrt{5}}{-2} when ± is plus. Add -8 to 2\sqrt{5}\approx 4.472135955.
x=4-\sqrt{5}
Divide -8+2\sqrt{5}\approx -3.527864045 by -2.
x=\frac{-2\sqrt{5}-8}{-2}
Now solve the equation x=\frac{-8±2\sqrt{5}}{-2} when ± is minus. Subtract 2\sqrt{5}\approx 4.472135955 from -8.
x=\sqrt{5}+4
Divide -8-2\sqrt{5}\approx -12.472135955 by -2.
-x^{2}+8x-11=-\left(x-\left(4-\sqrt{5}\right)\right)\left(x-\left(\sqrt{5}+4\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4-\sqrt{5}\approx 1.763932023 for x_{1} and 4+\sqrt{5}\approx 6.236067977 for x_{2}.
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