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x cubed - x squared + 8x - Factor Polynomial

by Zachary
(CA, USA)











































How do i factor the polynomial

P(x) = x³ - x² + 8x

for this problem, begin by finding the common factor in all three terms

factor out this common factor

the result will be a polynomial which is much easier to work with

Comments for x cubed - x squared + 8x - Factor Polynomial

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May 21, 2012
Factor x³ - x² + 8x
by: Staff



Question:

by Zachary
(CA, USA)

How do i factor a polynomial?


Answer:

P(x) = x³ - x² + 8x

P(x) = x*(x² - x + 8)


This equation cannot be factored beyond this point, unless you are willing to use complex numbers.


>>> the final answer is:

P(x) = x*(x² - x + 8)


However, “if” you are willing to use complex numbers (where i = √(-1) , you can calculate the other two roots using the quadratic formula:

ax² + bx + c = 0

x = [-b ± √(b² - 4ac)]/(2a)


x² - x + 8 = 0

x = unknown

a = +1

b = -1

c = +8


x = {-(-1) ± √[(-1)² - 4*1*(8)]}/(2*1)

x = {1 ± √(1 - 32)}/2

x = {1 ± √(-31)}/2


x = 1/2 ± √(-31)/2

x = 1/2 ± i√(31)/2

x = 0.5 ± 0.5i√(31)

x₁ = 0.5 + 0.5i√(31)

x₁ ≈ 0.5 + 0.5*i*5.56776436283

x₁ ≈ 0.5 + 2.783882181415*i



x₂ = 0.5 - 0.5i√(31)

x₂ ≈ 0.5 - 0.5*i*5.56776436283

x₂ ≈ 0.5 - 2.783882181415*i



When complex numbers (where i = √(-1) are used, the final answer is:


P(x) = x*[x - 0.5 + 0.5i√(31)]*[x - 0.5 - 0.5i√(31)]

or

P(x) = x * (x - 0.5 + 2.783882181415*i) * (x - 0.5 - 2.783882181415*i)


----------------------------------------------------------------------------------

Check the answer by multiplying the three factors together:

P(x) = x*[x - 0.5 + 0.5i√(31)]*[x - 0.5 - 0.5i√(31)]

P(x) = x*{[x - 0.5 + 0.5i√(31)]*[x - 0.5 - 0.5i√(31)]}

P(x) = x*{x² - 2*x*(0.5) + (0.5)² - [0.5i√(31)]²}

P(x) = x*{x² - x + 0.25 – (0.5)²*i²*[√(31)]²}

P(x) = x*{x² - x + 0.25 – (0.25)*i²*(31)}

P(x) = x*{x² - x + 0.25 – (7.75)*i²}

P(x) = x*{x² - x + 0.25 – (7.75)*[√(-1)]²}

P(x) = x*{x² - x + 0.25 – (7.75)*(-1)}

P(x) = x*{x² - x + 0.25 + 7.75}

P(x) = x*{x² - x + 8}

P(x) = x³ - x² + 8x, OK → the three factors are valid



Thanks for writing.

Staff
www.solving-math-problems.com



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