Degrees 4.0.1

Algebra -> Polynomials-and-rational-expressions-> SOLUTION: Form a polynomial whosezeros and degrees are given. Zeros:-3,-2,3; degree 3 A)f(x)=x^3-2x^2-9x+18 for a= 1 B)f(x)=x^3+2x^2-9x-18 for a= 1 C)f(x)=x^3-2x^2+9x-18 for a= 1 D) Log On
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Question 601396: Form a polynomial whosezeros and degrees are given.
Zeros:-3,-2,3; degree 3
A)f(x)=x^3-2x^2-9x+18 for a= 1
B)f(x)=x^3+2x^2-9x-18 for a= 1
C)f(x)=x^3-2x^2+9x-18 for a= 1
D)f(x)=x^3+2x^2+9x+18 for a= 1
Found 2 solutions by Alan3354, w_parminder:Answer by Alan3354(65904) (Show Source):
You can put this solution on YOUR website!
Form a polynomial whosezeros and degrees are given.
Zeros:-3,-2,3; degree 3
----------
Each zero, eg, z, contributes a factor (x - z)
Multiply (x+3)*(x-3)*(x+2)
---------------
=
=
B
----
I don't know what the a= 1 means.

Answer by w_parminder(53) (Show Source):
You can put this solution on YOUR website!
Sorry dear, I can't understand the problem completely but I am solving it to form a polynomial whose zeros are -3, -2 & +3 with degree 3
The simple method to find the polynomial is just write the factors of it and multiply them.
In order to find the factors, just subtract the zeros separately from a variable say 'X'
So the factors of the polynomial are
X - (-3) i.e. (X + 3)
X - (-2) i.e. (X + 2) &
X - (+3) i.e. (X - 3)
on multiplying these factors, we will get the required polynomial,
i.e. (X + 3)(X + 2)(X - 3)
i.e.
Now I don't know what is A, B, C & D
If these are the options for the correct answer then what is 'a = 1'
Please reply


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1 Answer

You can put this solution on YOUR website! Sorry dear, I can't understand the problem completely but I am solving it to form a polynomial whose zeros are -3, -2 & +3 with degree 3.

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Nov 15, 2016

Explanation:

A polynomial has #alpha# as a zero if and only if #(x-alpha)# is a factor of the polynomial. Working backwards, then, we can generate a polynomial with any zeros we desire by multiplying such factors.

Adobe audition cc 2019 v12.1.0.182. We want a polynomial #P(x)# with zeros #-3, 0, 1#, so:

#P(x) = (x-(-3))(x-0)(x-1)#

#=(x+3)x(x-1)#

Degrees 4.0.1 7

#=x(x+3)(x-1)#

#=x(x^2+2x-3)#

#=x^3+2x^2-3x#

Note that we could also multiply by any nonzero constant without changing the zeros, if a different polynomial is desired.

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