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blsea [12.9K]
3 years ago
15

What are the real zeros of x^3+4x^2-9x-36

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

x = -4\\x = -3\\x = 3

Step-by-step explanation:

To find the real zeros you must match the function to zero and factor the expression.

We have a polynomial of degree 3.

We try to group the terms to perform the factorization

x ^ 3 + 4x ^ 2-9x-36 = 0\\\\x ^ 2(x + 4) - 9(x + 4) = 0

Now we take (x + 4) as a common factor

(x + 4)(x ^ 2 -9) = 0

If we have an expression of the form (a ^ 2-b ^ 2) we know that this expression is equivalent to:

(a ^ 2-b ^ 2) = (a + b)(a-b)

In this case

a = x\\b = 3

So:

(x ^ 2 -9) = (x + 3)(x-3)

Finally:

x ^ 3 + 4x ^ 2-9x-36 = (x + 4)(x + 3)(x-3) = 0

The solutions are:

x = -4\\x = -3\\x = 3

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In this case, the answer choice that correctly calculates and interprets the standard deviation of the sum, S = X + Y is the sigma Subscript s = 0.8; companies A and B can expect the total weight of packages to vary by approximately 0.8 ounces from the mean.

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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
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1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

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188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

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Then,

188-5\times6^{2}

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So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

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Then,

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2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

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So, the number 1052, written as a base 6 number is 4512.

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