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

Beginner2021-12-27Added 44 answers

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:

$\Rightarrow f\left(x\right)=A{(x+2)}^{2}{(x-2)}^{3}$

When, x=0, f(0)=-108:

$\Rightarrow -108=A{(0+2)}^{2}{(0-2)}^{3}$

$\Rightarrow -108=A{\left(2\right)}^{2}{(-2)}^{3}$

$\Rightarrow -108=A\left(4\right)(-8)$

$\Rightarrow -108=-32A$

Step 3

$\Rightarrow \frac{-108}{-32}=A$

$\Rightarrow 3.375=A$

$\Rightarrow A=3.375$

$\Rightarrow f\left(x\right)=3.375{(x+2)}^{2}{(x-2)}^{3}$

Archie Jones

Beginner2021-12-28Added 34 answers

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, ${(x+2)}^{2}=0$

${(x-3)}^{3}=0$

or $f\left(x\right)=A{(x+2)}^{2}{(x-3)}^{3}$

f(0)=A(4)(-27)

-108=A(-108)

A=1

$\therefore f\left(x\right)={(x+2)}^{2}{(x-3)}^{3}$

Is the polynomial with degree of 5.

Vasquez

Expert2022-01-09Added 669 answers

Solution: zeros of -2 (multiplicity 2)

zeros of 3 (multiplicity 2)

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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