Math / Algebra

Questionand thoughts here and
Another way of thinking about this is to look at the output values ior values will all have the same sign (positive or negative).
3. Given the polynomial function $p(x)=(x-2)(x-5)^{4}(x+7)$ , what are all intervals on which $p(x) \leq 0$ ?
Complex Zeros A polynomial of degree $n$ has exactly $n$ complex zeros when counting multiplicities. "Complex" refers to both real and non-real zeros. If there are any non-real zeros, they always come in conjugate pairs (see explanation below). This means there are either no non-real zeros, or an even amount of non-real zeros. \begin{tabular}{|c|c|c|c|c|c|} \hline \multicolumn{3}{|l|}{Quadratic (degree $=2)$ } & \multicolumn{3}{|c|}{Cubic (degree $=3$ )} \\ \hline \# of real 0's & \# of real 0's & \# of real 0's & & \# of real 0's & \\ \hline \# of non-real 0 's & \# of non-real 0's & \# of non-real 0 's & \# of non-real 0 's & \# of non-real 0's & \# of non-real 0 's \\ \hline \end{tabular}
4. The degree of a polynomial is 8 with real zeros at $x=-10, x=5$ , and $x=16 . x=5$ has a multiplicity of 2 . How many non-real zeros does the polynomial have?
Non-real Zeros If $a+b i$ is a non-real zero of a polynomial $p$ , then its conjugate $\qquad$ is also a zero of $p$ .
Given one non-real zero of a polynomial, find another zero.
5. $-3+6 i$
6. $4-2 i$

Studdy Solution

STEP 1

1. The polynomial function is $p(x) = (x-2)(x-5)^4(x+7)$ .
2. We need to find the intervals where $p(x) \leq 0$ .

STEP 2

1. Identify the roots of the polynomial.
2. Determine the sign of $p(x)$ in the intervals defined by these roots.
3. Identify the intervals where $p(x) \leq 0$ .

STEP 3

Identify the roots of the polynomial $p(x) = (x-2)(x-5)^4(x+7)$ .
The roots are $x = 2$ , $x = 5$ , and $x = -7$ .

STEP 4

Determine the sign of $p(x)$ in the intervals defined by the roots: $(-\infty, -7)$ , $(-7, 2)$ , $(2, 5)$ , and $(5, \infty)$ .
- For $x < -7$ , all factors are negative, so $p(x) < 0$ . - For $-7 < x < 2$ , the factor $(x+7)$ is positive, others are negative, so $p(x) > 0$ . - For $2 < x < 5$ , the factors $(x-2)$ and $(x+7)$ are positive, $(x-5)^4$ is always positive, so $p(x) > 0$ . - For $x > 5$ , all factors are positive, so $p(x) > 0$ .

STEP 5

Consider the behavior at the roots: - At $x = -7$ , $p(x) = 0$ . - At $x = 2$ , $p(x) = 0$ . - At $x = 5$ , $p(x) = 0$ due to the even multiplicity of 4, $p(x)$ does not change sign.

STEP 6

Identify the intervals where $p(x) \leq 0$ .
The intervals are $(-\infty, -7]$ and $[5, \infty)$ .
The solution is that $p(x) \leq 0$ on the intervals $(-\infty, -7] \cup [5, \infty)$ .

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Studdy Solution

STEP 1

1. The polynomial function is $p(x) = (x-2)(x-5)^4(x+7)$ .
2. We need to find the intervals where $p(x) \leq 0$ .

STEP 2

1. Identify the roots of the polynomial.
2. Determine the sign of $p(x)$ in the intervals defined by these roots.
3. Identify the intervals where $p(x) \leq 0$ .

STEP 3

Identify the roots of the polynomial $p(x) = (x-2)(x-5)^4(x+7)$ .
The roots are $x = 2$ , $x = 5$ , and $x = -7$ .

STEP 4

STEP 5

Consider the behavior at the roots: - At $x = -7$ , $p(x) = 0$ . - At $x = 2$ , $p(x) = 0$ . - At $x = 5$ , $p(x) = 0$ due to the even multiplicity of 4, $p(x)$ does not change sign.

STEP 6

Identify the intervals where $p(x) \leq 0$ .
The intervals are $(-\infty, -7]$ and $[5, \infty)$ .
The solution is that $p(x) \leq 0$ on the intervals $(-\infty, -7] \cup [5, \infty)$ .

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Studdy Solution

STEP 1

1. The polynomial function is p(x)=(x−2)(x−5)4(x+7) p(x) = (x-2)(x-5)^4(x+7) p(x)=(x−2)(x−5)4(x+7).2. We need to find the intervals where p(x)≤0 p(x) \leq 0 p(x)≤0.

STEP 2

1. Identify the roots of the polynomial.2. Determine the sign of p(x) p(x) p(x) in the intervals defined by these roots.3. Identify the intervals where p(x)≤0 p(x) \leq 0 p(x)≤0.

STEP 3

Identify the roots of the polynomial p(x)=(x−2)(x−5)4(x+7) p(x) = (x-2)(x-5)^4(x+7) p(x)=(x−2)(x−5)4(x+7).The roots are x=2 x = 2 x=2, x=5 x = 5 x=5, and x=−7 x = -7 x=−7.

STEP 4

STEP 5

Consider the behavior at the roots: - At x=−7 x = -7 x=−7, p(x)=0 p(x) = 0 p(x)=0. - At x=2 x = 2 x=2, p(x)=0 p(x) = 0 p(x)=0. - At x=5 x = 5 x=5, p(x)=0 p(x) = 0 p(x)=0 due to the even multiplicity of 4, p(x) p(x) p(x) does not change sign.

STEP 6

Identify the intervals where p(x)≤0 p(x) \leq 0 p(x)≤0.The intervals are (−∞,−7] (-\infty, -7] (−∞,−7] and [5,∞) [5, \infty) [5,∞). The solution is that p(x)≤0 p(x) \leq 0 p(x)≤0 on the intervals (−∞,−7]∪[5,∞) (-\infty, -7] \cup [5, \infty) (−∞,−7]∪[5,∞).

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1. The polynomial function is $p(x) = (x-2)(x-5)^4(x+7)$ .
2. We need to find the intervals where $p(x) \leq 0$ .

1. Identify the roots of the polynomial.
2. Determine the sign of $p(x)$ in the intervals defined by these roots.
3. Identify the intervals where $p(x) \leq 0$ .

Identify the roots of the polynomial $p(x) = (x-2)(x-5)^4(x+7)$ .
The roots are $x = 2$ , $x = 5$ , and $x = -7$ .

Consider the behavior at the roots: - At $x = -7$ , $p(x) = 0$ . - At $x = 2$ , $p(x) = 0$ . - At $x = 5$ , $p(x) = 0$ due to the even multiplicity of 4, $p(x)$ does not change sign.

Identify the intervals where $p(x) \leq 0$ .
The intervals are $(-\infty, -7]$ and $[5, \infty)$ .
The solution is that $p(x) \leq 0$ on the intervals $(-\infty, -7] \cup [5, \infty)$ .