Mathematics, 21.10.2020 05:01 Val03

Which is a possible combination of root types for a 7th degree polynomial? 3 irrational roots, 4 complex roots 6 rational roots, 1 irrational root 7 rational roots 2 rational roots, 2 irrational roots, 3 complex roots

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Answer from: jaymariepope3334

There are four possible combinations of root types for a 7th degree polynomial:

(i) 7 rational roots. ( $a_{\mathbb{Q}} = 7$ , $a_{\mathbb{I}} = 0$ , $a_{\mathbb{C}} = 0$ )

(ii) 6 rational roots, 1 irrational root. ( $a_{\mathbb{Q}} = 6$ , $a_{\mathbb{I}} = 1$ , $a_{\mathbb{C}} = 0$ )

(iii) 2 rational roots, 1 irrational root, 4 complex roots. ( $a_{\mathbb{Q}} = 2$ , $a_{\mathbb{I}} = 1$ , $a_{\mathbb{C}} = 2$ )

(iv) 3 irrational roots, 4 complex roots. ( $a_{\mathbb{Q}} = 0$ , $a_{\mathbb{I}} = 3$ , $a_{\mathbb{C}} = 2$ )

Step-by-step explanation:

According to the Fundamental Theorem of Algebra, every polynomial with degree higher than zero has at least a solution and at best complex roots, where is the degree of the polynomial. In this case, we have 7 roots.

We can rewrite the 7th degree polynomial as a product of a 3rd degree polynomial and 4th degree polynomial. That is:

$p_{7}(x) = p_{3}(x)\cdot p_{4}(x)$ (Eq. 1)

From Algebra we know that both 3rd degree and 4th degree polynomials are solvable by analytical means:

1) All 4th degree polynomials are solvable by Ferrari's method, whose posible solutions are, respectively:

(i) 4 rational roots.

(ii) 4 irrational roots.

(iii) 2 rational roots, 2 irrational roots.

(iv) 3 rational roots, 1 irrational root.

(v) 1 rational roots, 3 irrational roots.

(vi) 2 rational roots, 2 complex roots.

(vii) 2 irrational roots, 2 complex roots.

(viii) 1 rational root, 1 irrational root, 2 complex roots.

(ix) 4 complex roots.

2) All 3rd degree polynomials are solvable by Cardano's method, whose possible solutions are, respectively:

(i) 3 rational roots.

(ii) 3 irrational roots.

(iii) 1 rational root, 2 complex roots.

(iv) 1 irrational root, 2 complex roots.

(v) 3 complex roots.

Please note that complex roots are always presented in conjugated pairs by Quadratic Formula.

The number of solutions of a polynomials are represented by the following expression:

$n = a_{\mathbb{Q}}+a_{\mathbb{I}}+2\cdot a_{\mathbb{C}}$ (Eq. 2)

Where:

$a_{\mathbb{Q}}$ - Number of rational roots, dimensionless.

$a_{\mathbb{I}}$ - Number of irrational roots, dimensionless.

$a_{\mathbb{C}}$ - Number of pairs of complex roots, dimensionless.

Based on this expression, we find the possible root types for a 7th degree polynomial:

(i) 7 rational roots. ( $a_{\mathbb{Q}} = 7$ , $a_{\mathbb{I}} = 0$ , $a_{\mathbb{C}} = 0$ )

(ii) 6 rational roots, 1 irrational root. ( $a_{\mathbb{Q}} = 6$ , $a_{\mathbb{I}} = 1$ , $a_{\mathbb{C}} = 0$ )

(iii) 2 rational roots, 1 irrational root, 4 complex roots. ( $a_{\mathbb{Q}} = 2$ , $a_{\mathbb{I}} = 1$ , $a_{\mathbb{C}} = 2$ )

(iv) 3 irrational roots, 4 complex roots. ( $a_{\mathbb{Q}} = 0$ , $a_{\mathbb{I}} = 3$ , $a_{\mathbb{C}} = 2$ )

Answer from: Quest

a net is a diagram that shows a “flattened” view of a solid. in a net, each face and base is shown with all of its dimensions. a net can also serve as a pattern to build a three-dimensional solid.

a pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex.

" the net of a triangular pyramid consists of four triangles. the base here we are given of the pyramid is a equilateral triangle, and the lateral faces are also triangles "

Answer from: Quest

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)

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