Let the GCD =x+1 and LCM = $x^{6} - 1$ for gives two polynomials. Then, if $f (x) = x^{3} + 1$ . what is g(x)?

(x^{3} + 1) (x - 1)

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(x^{3} - 1) (x + 1)

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x^{2} + 1 + x^{3}

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Solution

The correct option is B $(x^{3} - 1) (x + 1)$

$f (x) = x^{3} + 1$
GCD =x+1
$L C M = x^{6} - 1$
We know that , LCM $\times$ GCD =f(x) $\times$ g(x)
$\Rightarrow g (x) = \frac{L C M \times G C D}{f (x)} = \frac{(x^{6} - 1) (x + 1)}{x^{3} + 1} = \frac{(x^{3} + 1) (x^{3} - 1) (x + 1)}{x^{3} + 1} = (x^{3} - 1) (x + 1)$
That is, $g (x) = (x^{3} - 1) (x + 1)$

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Similar questions

Let GCD $= x + 1$ and LCM $= x^{6} - 1$ for two given polynomials $f (x)$ and $g (x)$ . Then, if $f (x) = x^{3} + 1$ . what is $g (x)$ ?

x^{3} + x^{2} - 1 = x^{2} + x^{3}

(x^{4} + 1) (x - x^{2})

x^{6} + x^{3} - x + 1

Let f (x) = x^{6} - 2 x^{5} + x^{3} + x^{2} - x - 1

g (x) = x^{4} - x^{3} - x^{2} - 1

Let a, b, c and d the roots of g (x) = 0

f (a) + f (b) + f (c) + f (d)

f : R \to R be f (x) = x^{3} + x^{2} + x - 1

g (x) = f^{- 1} (x)

g^{'} (2)

f (x) = 1 + x + x^{2} + x^{3} + \dots \dots

| x | < 1

f^{- 1} (x) =

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