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2 years ago

40x^-4 y^-6/(over)10x^-5 y^5

Mathematics

1 answer:

lisov135 [29]2 years ago

7 0

Answer:

4x/y¹¹

Step-by-step explanation:

1) take out constants

40/10 * x⁻⁴y⁻⁶/x⁻⁵y⁵

2) simplify 40/10 to 4

4 * x⁻⁴y⁻⁶/x⁻⁵y⁵

3) Use Quotient Rule: xᵃ/xᵇ=xᵃ⁻ᵇ

4x⁻⁴⁺⁵y⁻⁶⁻⁵

4) Simplify -4 + 5 to 1

4x¹y⁻⁶⁻⁵

5) simplify -6-5 to -11

4x¹y⁻¹¹

6) use rule of one: ₓ¹=ₓ

4xy⁻¹¹

7) use negative power rule: x⁻ᵃ=1/xᵃ

4x * 1/y¹¹

8) simplify

4x/y¹¹

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The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3

<h3>How to determine the number of real zeros?</h3>

The equation of the function is given as:

$f(x) = x^3 + 4x^2 + x - 6$

Expand the function

$f(x) = x^3 + 5x^2 - x^2 + 6x - 5x - 6$

Reorder the terms

$f(x) = x^3 + 5x^2 + 6x - x^2 - 5x - 6$

Factor the expression

$f(x) = x(x^2 + 5x + 6) -1(x^2 + 5x + 6)$

Factor out x -1

$f(x) = (x^2 + 5x + 6)(x -1)$

Expand

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

Factorize

$f(x) = [x(x + 3) + 2(x + 3)](x - 1)$

Factor out x + 2

$f(x) = (x + 3)(x + 2)(x- 1)$

The function has been completely factored and it has 3 linear factors

Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3