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PLEASE HELP!!!
Form a polynomial​ f(x) with real coefficients having the given degree and zeros.
Degree​ 4; ​ zeros: 6​, multiplicity​ 2; 4i

Enter the polynomial. Let a represent the leading coefficient.
​f(x)=a(__)

PLEASE HELP Form A Polynomial Fx With Real Coefficients Having The Given Degree And Zeros Degree 4 Zeros 6 Multiplicity 2 4i Enter The Polynomial Let A Represen class=

Sagot :

briannaseptember18

Answer:

f(x)=a(x4−12x3+52x2−192x+576)

Step-by-step explanation:

The polynomial can be expressed as:f(x)=a(x−6)2(x−4i)(x+4i)f(x)=a(x−6)2(x−4i)(x+4i)First, handle the complex zeros:(x−4i)(x+4i)=x2−(4i)2=x2−(−16)=x2+16(x−4i)(x+4i)=x2−(4i)2=x2−(−16)=x2+16Thus, the polynomial is:f(x)=a(x−6)2(x2+16)f(x)=a(x−6)2(x2+16)Now, expand (x−6)2(x−6)2:(x−6)2=x2−12x+36(x−6)2=x2−12x+36Substituting back, we get:f(x)=a(x2−12x+36)(x2+16)f(x)=a(x2−12x+36)(x2+16)Expand the product:(x2−12x+36)(x2+16)=x2(x2+16)−12x(x2+16)+36(x2+16)=x4+16x2−12x3−192x+36x2+576(x2−12x+36)(x2+16)=x2(x2+16)−12x(x2+16)+36(x2+16)=x4+16x2−12x3−192x+36x2+576Combine like terms:f(x)=a(x4−12x3+52x2−192x+576)f(x)=a(x4−12x3+52x2−192x+576)So, the polynomial with the given conditions is:f(x)=a(x4−12x3+52x2−192x+576)f(x)=a(x4−12x3+52x2−192x+576)Thus, the polynomial f(x)f(x) in the form requested is:f(x)=a(x4−12x3+52x2−192x+576)f(x)=a(x4−12x3+52x2−192x+576)

semsee45

Answer:

[tex]\textsf{1)}\quad f(x)=a(x^4-12x^3+52x^2-192x+576)[/tex]

[tex]\textsf{2)}\quad 8+i[/tex]

Step-by-step explanation:

Question 1

To form a polynomial f(x) with real coefficients given its zeros, we can use the factored form of a polynomial:

[tex]\boxed{\begin{array}{l}\underline{\textsf{Factored form of a polynomial}}\\\\y=a(x-r_1)(x-r_2)...(x-r_n)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a$ is the leading coefficient}.\\\phantom{ww}\bullet\;\textsf{$r_1, r_2, ..., r_n$ are the zeros (roots)}.\\\end{array}}[/tex]

The number of factors corresponds with the degree of the polynomial.

Given conditions:

  • Polynomial function f(x) with real integer coefficients.
  • Degree: 4
  • Zeros: 6 with multiplicity of 2; 4i

If a polynomial with real coefficients has a complex root (a + bi), then its conjugate (a - bi) is also a root (zero) of the polynomial.

So, given that 4i is a zero of function f(x), its complex conjugate -4i is also a zero of the polynomial.

In the context of polynomial roots, the multiplicity refers to how many times a particular root appears as a factor in the factored form of the polynomial. Therefore, given that the root 6 has a multiplicity of 2, it appears twice.

So, the factored form of polynomial f(x) is:

[tex]f(x)=a(x-6)(x-6)(x-4i)(x-(-4i))[/tex]

Expand and simplify:

[tex]f(x)=a(x^2-12x+36)(x-4i)(x+4i) \\\\f(x)=a(x^2-12x+36)(x^2+4ix-4ix-16i^2) \\\\f(x)=a(x^2-12x+36)(x^2-16(-1)) \\\\f(x)=a(x^2-12x+36)(x^2+16) \\\\f(x)=a(x^4+16x^2-12x^3-192x+36x^2+576) \\\\f(x)=a(x^4-12x^3+52x^2-192x+576)[/tex]

Therefore, the polynomial f(x) with real coefficients having the given degree and zeros is:

[tex]\Large\boxed{\boxed{f(x)=a(x^4-12x^3+52x^2-192x+576)}}[/tex]

[tex]\dotfill[/tex]

Question 2

The number of zeros of a polynomial is equal to its degree, counting multiplicities and including complex conjugate pairs.

Therefore, given that the degree of polynomial f(x) is 3 and we are provided with two zeros, 3 and (8 - i), there is one remaining zero to find.

Since the coefficients of f(x) are real numbers, complex roots occur in conjugate pairs. Therefore, if 8 - i is a root, its conjugate 8 + i must also be a root of f(x).

So, the remaining zero of f is:

[tex]\Large\boxed{\boxed{8+i}}[/tex]

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