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olga2289 [7]
3 years ago
11

Explain how to find the zeros of the given polynomial: x3+3x2–x−3 What are the zeros?

Mathematics
2 answers:
guapka [62]3 years ago
4 0

Answer:

The zeroes of the function are x = -3, x = -1, and x = 1.

Step-by-step explanation:

We are given a polynomial and asked to find the zeroes of the polynomial. Our given polynomial is:

x^3 + 3x^2 - x - 3

The zeroes of a polynomial are the solutions or roots of the function. This is where if the line were graphed, the line would intersect the x-axis at those points (sometimes there will only be one).

This is a cubic polynomial. We can use a factoring technique referred to as grouping, which is most effective when factoring a trinomial with the only satisfaction met is: a > 1.

However, it can also be used when a > 1 is not true.

We can work with an example polynomial first and then work through solving the given polynomial.

An example polynomial would be:

10x^2 + 8x - 9 = 0

In grouping, our main goal is to factor as quickly as possible. If you examine the polynomial, there is no common factor between 10x², 8x, or -9, so we cannot take out a GCF and need to use another factoring technique.

Since a > 1, we can instantly jump to using the grouping factoring technique.

<em>Please note that the parent function for a quadratic is </em>ax^2 + bx + c = 0.

To group:

  1. We want to find the product of the terms a and c. So, with our example polynomial, a is 10, b is 8, and c is -9. Therefore, let's find the product of 10 and -9.
  2. Using ac, we will list all available factors. After doing this, we will find two factors of ac that will add up to give us b.
  3. These two factors are then substituted with b and c in the following parent equation: ax^2 + bx + cx + d.
  4. Finally, the equation is factored using parentheses in the following format: (ax^2 + bx)(+ cx + d) = 0
  5. After we group the function, we take the GCF out of both parentheses sets and write it as a coefficient to the remaining bit of the parentheses. The two GCFs are a set of terms and the remaining portion is the other set of terms.

Let's use grouping to solve x^3 + 3x^2 - x - 3.

Because we already have four terms, we can skip steps one through three and go straight to step four.

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

Now, we need to take the GCF of both parentheses sets. Let's do this with our first set of terms.

  • You can divide x² out of both x³ and 3x², so x² is the GCF, leaving x and 3 behind.

Then, we can also do this with our second set of terms.

  • There is a -1 implied in front of x, and there is no other common term between -x and -3, so -1 is our GCF, leaving x and 3 behind.

Because our remaining terms for both sets match, we have grouped our polynomial correctly. Now, let's combine our GCFs into one term and list our common term as the second term.

(x^2-1)(x+3)

Now, we can break apart x² - 1 into more factors.

This is the difference of two squares. The formula for the difference of two squares is (a^2-b^2)=(a+b)(a-b)

Our perfect squares are every integer in the spectrum of numbers, so 1 is a perfect square. Therefore, x² becomes x and 1 becomes ±1.

Our new terms would therefore be (x+1)(x-1)(x+3).

Finally, to find zeroes of a function, x needs to equal <em>something</em>. Therefore, we can set our terms equal to zero and solve for x.

<u>Term 1 (x + 1)</u>

x + 1 = 0\\\\x + 1 - 1 = 0 - 1\\\\\boxed{x = -1}

<u>Term 2 (x - 1)</u>

<u />x - 1 = 0\\\\x - 1 + 1 = 0 + 1\\\\\boxed{x = 1}<u />

<u>Term 3 (x + 3)</u>

<u />x + 3 = 0\\\\x + 3 - 3 = 0 - 3\\\\\boxed{x = -3}<u />

Therefore, our zeroes are listed above. To finalize our answer, we want to list them in increasing order, so this will be most negative to most positive. Therefore, our final answer is x = -3, -1, 1.

AveGali [126]3 years ago
3 0

Answer:

Step-by-step explanation:

The zero (root) of a function is any value of the variable that will produce an answer of zero. Graphically, the real zero of a function is where the graph of the function crosses the x-axis.

~~~~~~~~~

f(x) = x³ + 3x² - x - 3

f(-3) = ( - 3 )³ + 3( - 3 )² - ( - 3 ) - 3 = 0

f(-1) = ( - 1 )³ + 3( - 1 )² - ( - 1 ) - 3 = 0

f(1) = ( 1 )² + 3( 1 )² - ( 1 ) - 3 = 0

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