Why is it possible to split (x-1)(x+1)(x-5) = 0 into 3 equations?

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It's easy to see if x x is a root, then so is −1/x. − 1 / x . That means x+1/x x + 1 / x is the difference of the roots.

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Given: x + 1/x = 5 Formula used: (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3ab(a + b) + b3 Calculation: x + 1/x = 5 ⇒ x2 + 1/x2 = 52 &nda.

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Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=n∑k=0nCk⋅(an−kbk) ( a + b ) n = ∑ k = 0 n ⁡ n C k ⋅ ( a n - k b k ) .

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x + 1/x = 5 implies (x^2/x) + 1/x = 5. Combining fractions on LHS gives (x^2 + 1)/x = 5/1. Cross multiplying gives us x^2 + 1

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X+1/X=5..1 (X+1/X)^2=5^2 X^2+1/X^2+2=25 X^2+1/X^2=23 {X^2+1/X^2}^2=23^2 X^4+1/X^4=529-2 X^4+1/(X^4)=527... 2 1*2 (X+1/X)(X^4+1/X^4)= 5*527 ...

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The Algebra Calculator provides step-by-step help for algebra problems. Type your problem into the text box, like 3x+2=14, to get a solution.

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Because anything multiplied by zero is zero, since you're multiplying between groups of parenthesis, if the value in any one is 0, the whole equation is 0.

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