MATHEMATICS MIDDLE SCHOOL

Find all the zeroes of the polynomial p(x) = x4 – 7x³ + 9x² + 13x – 4, if two of its zeroes are (2 +√ 3) and ( 2 - √ 3).

Answers

Answer 1

Answer:

All the zeroes of the polynomial p(x) are:

$-1\ ,\ 4\ ,\ (2+√(3))\ ,\ (2-√(3))$

Step-by-step explanation:

We are given a polynomial expression p(x) as:

$p(x)=x^4-7x^3+9x^2+13x-4$

Also two of the zeros of the polynomial are:

$2+√(3)\ and\ 2-√(3)$

i.e. the polynomial could be factored as follows:

$x^4-7x^3+9x^2+13x-4=(x-(2+√(3)))(x-(2-√(3)))q(x)$

where q(x) is a polynomial of degree 2

i.e.

$x^4-7x^3+9x^2+13x-4=(x-(2+√(3)))(x-(2-√(3)))q(x)$

i.e.

$q(x)=(x^4-7x^3+9x^2+13x-4)/(x^2-4x+1)$

Hence, we get the value of q(x) as:

$q(x)=x^2-3x-4\n\n\ni.e.\n\n\nq(x)=x^2-4x+x-4\n\n\ni.e.\n\n\nq(x)=x(x-4)+1(x-4)\n\n\ni.e.\n\n\nq(x)=(x+1)(x-4)$

Hence, our polynomial could be represented in terms of linear factors as:

$x^4-7x^3+9x^2+13x-4=(x-(2+√(3)))(x-(2-√(3)))(x+1)(x-4)$

i.e. the other two zeros are: -1 and 4

Answer 2

Answer: polynomial of the two zeroes
(x - (2+rt(3))(x - (2-rt(3)) = x^2 -4x +1

divide it
(x^4 - 7x^3 + 9x^2 + 13x – 4) / (x^2 -4x +1)
= x^2 - 3x -4
= (x-4)(x+1)
x=-1, 4 ... the remaining zeroes

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Answer:

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Step-by-step explanation:

the degree measure of other would be 15 degrees

because if you add all the other ones up that will be 85 and the rest is 15

so the degree would be 54

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