First, we need to find which number when substituted into the equation will give the answer zero. \(f(1) = {(1)^3} + 4{(1)^2} + (1) - 6 = 0\) Therefore \((x - 1)\)is a factor. Factorise the quadratic ...
This is a preview. Log in through your library . Abstract Let $\{\lambda _{j}\}_{j=0}^{\infty}$ be a sequence of distinct positive numbers. Let 1 ≤ p ≤ ∞ and ...
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