Question

Determine how many, what type, and find the roots for f(x) = x4 + 21x2 − 100. Determine how many, what type, and find the roots for f(x) = x3 − 5x2 − 25x + 125. The following graph shows a seventh-degree polynomial: graph of a polynomial that touches the x axis at negative 5, crosses the x axis at negative 1, crosses the y axis at negative 2, crosses the x axis at 4, and crosses the x axis at 7. Part 1: List the polynomial’s zeroes with possible multiplicities. Part 2: Write a possible factored form of the seventh degree function. Without plotting any points other than intercepts, draw a possible graph of the following polynomial: f(x) = (x + 8)3(x + 6)2(x + 2)(x − 1)3(x − 3)4(x − 6).

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

For the polynomial, $f(x)=x^4+21x^2-100$, the roots are $2,-2,j5,-j5$.

For the polynomial, $f(x)=x^3-5x^2-25x+125$, the roots are $5;5;-5$.

For the polynomial, $f(x)=x^4+21x^2-100$, the degree is $4$ so, the polynomial is a quartic polynomial. The total number of roots are also $4$. To find the roots of the polynomial, factorise it as

$x^4+21x^2-100=0\\(x^2-4)(x^2+25)=0\\(x-2)(x+2)(x-j5)(x+j5)=0\\x=2;-2;j5;-j5$

For the polynomial, $f(x)=x^3-5x^2-25x+125$, the degree is $3$ so, the polynomial is a cubic polynomial. The total number of roots are also $3$. To find the roots of the polynomial, factorise it as

$x^3-5x^2-25x+125=0\\(x-5)(x^2-25)=0\\(x-5)(x-5)(x+5)=0\\x=5;5;-5$

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