Third-degree, with zeros of -3,-2, and 1, and passes through the point (3,9)

Answers

Answer 1

Three zeroes of a third-degree polynomial are given, with the graph passing through the point (3,6). To find the equation of this, first, let us set up how a 3rd-degree polynomial factored out should look like:  

y =  a(x+b)(x+c)(x+d).

Since the zeroes of the function are -3, -2, 1, the equation will look like y = a(x+3)(x+2)(x-1). To solve for a, plug-in x = 3 and y = 6 from the point (3,6). After replacing the variable a in the equation with the answer, you will have the solution for the problem.

Answer 2

Third-degree, with zeros of -3,-2, and 1, and passes through the point (3,9) then the polynomial p(x) is (3/20)(x-1)(x+2)(x+3)

What is Polynomial?

Polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and positive-integer powers of variables

-3 is a root... (x+3) is a factor

-2 is a root... (x+2) is a factor

1 is a root... (x-1) is a factor

p(x) = a(x-1)(x+2)(x+3)

(3, 9) is on graph so p(3)=9

9=a(3-1)(3+2)(3+3)

9=a(2)(5)(6)

9=60a

Divide both sides by 60

a=9/60

a=3/20

p(x) = (3/20)(x-1)(x+2)(x+3)

Hence,  third-degree, with zeros of -3,-2, and 1, and passes through the point (3,9) then the polynomial p(x) is (3/20)(x-1)(x+2)(x+3)

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Answer:

hope it helps you............

[tex]\bold{Hewo!!}[/tex]

--------------------------

ANSWER :[tex]\sf{=(y~+~1)~(y-2)[/tex]EXPLANATION :

[tex]\sf{Factor~y^2 - y-2:}~(y~+~1)~(y-2)[/tex]

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[tex]\underline{\large\huge{\mathfrak{Answer~:}}}[/tex]

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Those are the steps that the calculator said to do ^

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