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How do you factor x to the 8th power minus 1?
x8-1 becomes (x4 + 1) (x4 - 1) - perfect square and then (x4 + 1) (x2 - 1)(x2 + 1) - perfect square again. and then (x4 + 1)(x2 + 1)(x + 1)(x - 1) - perfect square again
X4 - 36 factor the polynomial in quadratics form?
-32
X4 plus 7x2 minus 60 factor completely?
x4 + 7x2 - 60 = x4 + 12x2 - 5x2 - 60 = x2(x2 + 12) - 5(x2 +12) = (x2 - 5)(x2 +12)
How do you factor when there's powers in the equation?
A power can be factored in many different ways; for example: x5 = x times x4 = x2 times x3. When such a power appears as a term in combination with other expressions, you should look out for common factoring patterns. Here are two examples: x2 + x5 Here, you can use the pattern "common factor". In this case, both parts have the common factor x2, so you can factor it out. x4 - 1 Here, you can use the pattern "difference of squares". Note that x4 is the square of x2, and 1 is the square of 1.
Is x-2 a factor of fx equals -9 x4 plus 3 x3 - 7 x2 plus 7 x - 9?
If there is a remainder then it's not a factor otherwise yes
