Question

Write a polynomial function with the given zeros. multiply the terms (write in standard form).

Answer 1

$1) \ f(x)=x^3-3x-2 \\ \\ 2) \ g(x)=x^3-x^2-6x$

Explanation:

$A \ \mathbf{polynomial \ function} \ of \ x \ with \ degree \ n \ is \ given \ by:\\ \\ f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots +a_{2}x^{2}+a_{1}x+a_{0} \\ \\ where \ n \ is \ a \ nonnegative \ integer \ and \ a_{n}, a_{n-1}, \ldots a_{2}, a_{1}, a_{0} \\ with \ a_{n}\neq 0$

In this exercise, we have two cases and both will be polynomial functions with degree 3 because we have three real roots in each case. So:

First.

The roots are:

$x= -1, -1, 2$

So we can write this polynomial functions as the product of linear factors:

$f(x)=(x-(-1))(x-(-1))(x-2) \\ \\ f(x)=(x+1)^2(x-2)$

Since we have to write it in standard form, let's expand:

$f(x)=(x+1)^2(x-2) \\ \\ \\ By \ property: \\ \\ (a+b)^2=a^2+2ab+b^2 \\ \\ \\ f(x)=(x^2+2x+1)(x-2) \\ \\ f(x)=x^3+2x^2+x-2x^2-4x-2 \\ \\ \\ Combine \ like \ terms: \\ \\ x^3+(2x^2-2x^2)+(x-4x)-2 \\ \\ \\ \boxed{f(x)=x^3-3x-2}$

Second.

The roots are:

$x= -2, 0, 3$

Writing the polynomial function as the product of linear factors:

$g(x)=(x-(-2))(x-0)(x-3) \\ \\ g(x)=(x+2)x(x-3) \\ \\ g(x)=x(x+2)(x-3) \\ \\ \\ Distributive \property: \\ \\ g(x)=x(x^2-3x+2x-6) \\ \\ g(x)=x(x^2-x-6) \\ \\ \\ Distributive \ property \ again: \\ \\ \boxed{g(x)=x^3-x^2-6x}$

Learn more:

Degree of polynomial functions: brainly.com/question/5451252

#LearnWithBrainly

Answer 2

1) polynomial having zeros x=-1,-1,2 is $x^3-3x-2=0$

2) polynomial having zeros x= -2, 0, 3 is $x^3-x^2-6x=0$

Step-by-step explanation:

1) We are given the zeros:

x=-1,-1,2

we need to find the polynomial

x=-1, x=-1, x=2

x+1=0,x+1=0, x-2=0

Multiplying all terms:

$(x+1)(x+1)(x-2)=0\\(x(x+1)+1(x+1))(x-2)=0\\(x^2+x+x+1)(x-2)=0\\(x^2+2x+1)(x-2)=0\\x(x^2+2x+1)-2(x^2+2x+1)=0\\x^3+2x^2+x-2x^2-4x-2=0\\x^3+0x^2-3x-2=0\\x^3-3x-2=0$

So, polynomial having zeros x=-1,-1,2 is $x^3-3x-2=0$

2) We are given the zeros:

x= -2, 0, 3

we need to find the polynomial

x= -2, x=0, x=3

x+2=0,x=0, x-3=0

Multiplying all terms:

$x(x+2)(x-3)=0\\x(x(x-3)+2(x-3))=0\\x(x^2-3x+2x-6)=0\\x(x^2-x-6)=0\\x^3-x^2-6x=0$

So, polynomial having zeros x= -2, 0, 3 is $x^3-x^2-6x=0$

Keywords: Zeros of polynomials

Learn more about zeros of polynomials at:

• brainly.com/question/1414350
• brainly.com/question/2568692
• brainly.com/question/1332667
• brainly.com/question/1357167

#learnwithBrainly