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Determine the equation of a cubic polynomial function with roots at (3,0), (2,0) and (-4,0) that also passes through the point (1,30).

Respuesta :

culveryolanda46

Step-by-step explanation:

Step 1: Setting Up the Factors

Our roots are 3, 2, and -4 so

our factors are

[tex](x - 3)(x - 2)(x + 4)[/tex]

Step 2: Initial Test to see if the result equal 30

Let a be a constant, such we have

[tex]a(x - 3)(x - 2)(x + 4) = 30[/tex]

Plug in. 1 for x.

[tex]a(1 - 3)(1 - 2)(1 + 4) = 30[/tex]

[tex]a( - 2)( - 1)(5) = 30[/tex]

[tex]a(10) = 30[/tex]

[tex]a = 3[/tex]

So our equation is

[tex]3(x - 3)(x - 2)(x + 4)[/tex]

Or if you want it simplifed

[tex]3( {x}^{2} - 5x + 6)(x + 4) = 3( {x}^{3} + - {x}^{2} - 14x + 24) = 3 {x}^{3} - 3 {x}^{2} - 42x + 72[/tex]

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semsee45

Answer:

[tex]f(x)=3(x-3)(x-2)(x+4)[/tex]

Step-by-step explanation:

General form of a cubic polynomial function with 3 roots:

[tex]f(x)=a(x-b)(x-c)(x-d)[/tex]

where:

  • a is some constant to be found
  • b, c and d are the roots of the function

Given roots:

  • (3, 0)
  • (2, 0)
  • (-4, 0)

Substitute the given roots into the general form of the function:

[tex]\implies f(x)=a(x-3)(x-2)(x-(-4))[/tex]

[tex]\implies f(x)=a(x-3)(x-2)(x+4)[/tex]

To find the value of a, substitute the given point (1, 30) into the equation:

[tex]\begin{aligned} f(1) & = 30\\\implies a(1-3)(1-2)(1+4) & =30\\a(-2)(-1)(5) & = 30 \\10a & = 30\\\implies a & = 3 \end{aligned}[/tex]

Therefore, the equation of the cubic polynomial function is:

[tex]f(x)=3(x-3)(x-2)(x+4)[/tex]

Learn more about polynomials here:

https://brainly.com/question/27953978

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