Given the polynomials $p(z)=z^{\wedge} 4-16 z^{\wedge} 2+40 z-25$ and $d(z)=z^{\wedge} 2-4 z+5$.
a) Perform division of $p(z)$ by $d(z)$.
b) Determine from the result in a) all the roots of the polynomial $p(z)$.

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the polynomials $p(z) = z^4 - 16z^2 + 40z - 25$ and $d(z) = z^2 - 4z + 5$, we need to perform the division of $p(z)$ by $d(z)$. ‖ ‖ step 2 ⋮ We start by expressing the division as $\frac{p(z)}{d(z)} = \frac{z^4 - 16z^2 + 40z - 25}{z^2 - 4z + 5}$. ‖ ‖ step 3 ⋮ Using polynomial long division or synthetic division, we find that $\frac{z^4 - 16z^2 + 40z - 25}{z^2 - 4z + 5} = z^2 + 4z - 5$. ‖ ‖ step 4 ⋮ Therefore, $p(z) = (z^2 - 4z + 5)(z^2 + 4z - 5)$. ‖ ‖ step 5 ⋮ To find the roots of $p(z)$, we solve the equation $p(z) = 0$. ‖ ‖ step 6 ⋮ This gives us $(z^2 - 4z + 5)(z^2 + 4z - 5) = 0$. ‖ ‖ step 7 ⋮ Solving $z^2 - 4z + 5 = 0$ using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $z = 2 \pm i$. ‖ ‖ step 8 ⋮ Solving $z^2 + 4z - 5 = 0$ using the quadratic formula, we get $z = -5$ and $z = 1$. ‖ ∻Answer∻ ⚹ The roots of the polynomial $p(z)$ are $z = -5$, $z = 1$, $z = 2 - i$, and $z = 2 + i$. ⚹ ∻Key Concept∻ ⚹Polynomial Division and Root Finding⚹ ∻Explanation∻ ⚹We performed polynomial division to factorize $p(z)$ and then solved the resulting quadratic equations to find all the roots.⚹◊What is the quotient obtained when dividing $p(z)$ by $d(z)$ in part (a)?⍭ Generate me a similar question◊

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