 b2sonicxh

2021-12-26

Write a polynomial f (x) that meets the given conditions and has integer coefficients.
Lowest degree polynomial with f(0) = -108 and zeros of -2 (multiplicity 2) and 3 (multiplicity 2) William Appel

Step 1
Given:
Polynomial of lowest degree with zeros of -2 (multiplicity 2) and 3(multiplicity 2)
and with f(0)=-108.
Step 2
The polynomial equation is provided by:
$⇒f\left(x\right)=A{\left(x+2\right)}^{2}{\left(x-2\right)}^{3}$
When, x=0, f(0)=-108:
$⇒-108=A{\left(0+2\right)}^{2}{\left(0-2\right)}^{3}$
$⇒-108=A{\left(2\right)}^{2}{\left(-2\right)}^{3}$
$⇒-108=A\left(4\right)\left(-8\right)$
$⇒-108=-32A$
Step 3
$⇒\frac{-108}{-32}=A$
$⇒3.375=A$
$⇒A=3.375$
$⇒f\left(x\right)=3.375{\left(x+2\right)}^{2}{\left(x-2\right)}^{3}$ Archie Jones

Step 1
To find a polynomial f (x) with integer coefficients that satisfies the requirements
Step 2
Polynomial of lowest-degree
with zeroes of -2 multiplicity 2...(i)
zeros of 3 multiplicity 3...(ii)
f(0)=-108
i*e, ${\left(x+2\right)}^{2}=0$
${\left(x-3\right)}^{3}=0$
or $f\left(x\right)=A{\left(x+2\right)}^{2}{\left(x-3\right)}^{3}$
f(0)=A(4)(-27)
-108=A(-108)
A=1
$\therefore f\left(x\right)={\left(x+2\right)}^{2}{\left(x-3\right)}^{3}$
Is the polynomial with degree of 5. Vasquez

Solution: zeros of -2 (multiplicity 2)
zeros of 3 (multiplicity 2)
$\begin{array}{}\mathrm{△}f\left(0\right)=-108\\ ⇒f\left(x\right)=k\left(x+2{\right)}^{2}\left(x-3{\right)}^{2}\\ f\left(0\right)=-108=k\left(4\right)\left(9\right)=36k\\ k=\frac{-108}{36}=-3\\ ⇒f\left(x\right)=-3\left(x+2{\right)}^{2}\left(x-3{\right)}^{2}\end{array}$

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