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Radda [10]
3 years ago
6

Huh.. can someone please help me, i honestly really need this rn.. :(

Mathematics
1 answer:
Harman [31]3 years ago
7 0

Answer:

If

€

p(x) is a polynomial, the solutions to the equation

€

p(x) = 0 are called the zeros of the

polynomial. Sometimes the zeros of a polynomial can be determined by factoring or by using the

Quadratic Formula, but frequently the zeros must be approximated. The real zeros of a polynomial

p(x) are the x-intercepts of the graph of

€

y = p(x).

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The Factor Theorem: If

€

(x − k) is a factor of a polynomial, then

€

x = k is a zero of the polynomial.

Conversely, if

€

x = k is a zero of a polynomial, then

€

(x − k) is a factor of the polynomial.

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Example 1: Find the zeros and x-intercepts of the graph of

€

p(x) =x

4−5x

2 + 4.

€

x

4−5x

2 + 4 = 0

(x

2 − 4)(x

2 −1) = 0

(x + 2)(x − 2)(x +1)(x −1) = 0

x + 2 = 0 or x − 2 = 0 or x +1= 0 or x −1= 0

x = −2 or x = 2 or x = −1 or x =1

So the zeros are –2, 2, –1, and 1 and the x-intercepts are (–2,0), (2,0), (–1,0), and (1,0).

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The number of times a factor occurs in a polynomial is called the multiplicity of the factor. The

corresponding zero is said to have the same multiplicity. For example, if the factor

€

(x − 3) occurs to

the fifth power in a polynomial, then

€

(x − 3) is said to be a factor of multiplicity 5 and the

corresponding zero, x=3, is said to have multiplicity 5. A factor or zero with multiplicity two is

sometimes said to be a double factor or a double zero. Similarly, a factor or zero with multiplicity

three is sometimes said to be a triple factor or a triple zero.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 2: Determine the equation, in factored form, of a polynomial

€

p(x) that has 5 as double

zero, –2 as a zero with multiplicity 1, and 0 as a zero with multiplicity 4.

€

p(x) = (x − 5)

2(x + 2)x

4

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Example 3: Give the zeros and their multiplicities for

€

p(x) = −12x

4 + 36x3 − 21x

2.

€

−12x

4 + 36x3 − 21x

2 = 0

−3x

2(4x

2 −12x + 7) = 0

−3x

2 = 0 or 4x

2 −12x + 7 = 0

x

2 = 0 or x = −(−12)± (−12)

2−4(4)(7)

2(4)

x = 0 or x = 12± 144−112

8 = 12± 32

8 = 12±4 2

8 = 12

8 ± 4 2

8 = 3

2 ± 2

2

So 0 is a zero with multiplicity 2,

€

x = 3

2 − 2

2 is a zero with multiplicity 1, and

€

x = 3

2 + 2

2 is a zero

with multiplicity 1.

(Thomason - Fall 2008)

Because the graph of a polynomial is connected, if the polynomial is positive at one value of x and

negative at another value of x, then there must be a zero of the polynomial between those two values

of x.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 4: Show that

€

p(x) = 2x3 − 5x

2 + 4 x − 7 must have a zero between

€

x =1 and

€

x = 2.

€

p(1) = 2(1)

3 − 5(1)

2 + 4(1) − 7 = 2(1) − 5(1) + 4 − 7 = 2 − 5 + 4 − 7 = −6

and

€

p(2) = 2(2)3 − 5(2)

2 + 4(2) − 7 = 2(8) − 5(2) + 8 − 7 =16 −10 + 8 − 7 = 7.

Because

€

p(1) is negative and

€

p(2) is positive and because the graph of

€

p(x) is connected,

€

p(x)

must equal 0 for a value of x between 1 and 2.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

If a factor of a polynomial occurs to an odd power, then the graph of the polynomial actually goes

across the x-axis at the corresponding x-intercept. An x-intercept of this type is sometimes called an

odd x-intercept. If a factor of a polynomial occurs to an even power, then the graph of the

polynomial "bounces" against the x-axis at the corresponding x-intercept, but not does not go across

the x-axis there. An x-intercept of this type is sometimes called an even x-intercept.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Example 5: Use a graphing calculator or a computer program to graph

€

y = 0.01x

2(x + 2)3(x − 2)(x − 4)

4 .

x

y

–2 2 4

5

Because the factors

€

(x + 2) and

€

(x − 2) appear to odd

powers, the graph crosses the x-axis at

€

x = −2

and

€

x = 2.

Because the factors x and

€

(x − 4) appear to even

powers, the graph bounces against the x-axis at

€

x = 0

and

€

x = 4.

Note that if the factors of the polynomial were

multipled out, the leading term would be

€

0.01x10.

This accounts for the fact that both tails of the graph

go up; in other words, as

€

x → −∞,

€

y

Step-by-step explanation:

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Y_Kistochka [10]
Well 2.5 + 2.5 is 5 so 5 - 2.5 is 2.5
5 0
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Select the correct answer.<br> Find the value of g(7) for the function below.<br> g(x) = zur
ale4655 [162]

Answer:

B

Step-by-step explanation:

To evaluate g(7) substitute x = 7 into g(x) , that is

g(7) = \frac{7}{8} × 7 - \frac{1}{2}

      = \frac{49}{8} - \frac{4}{8}

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5 0
3 years ago
Hasil dari 64^5/6 + 27^2/3 - 81^3/4 adalah<br>tolong bantuannya kakak kakak<br>​
NeTakaya

Answer:

14

Step-by-step explanation:

Given the indicinal expression 64^5/6 + 27^2/3 - 81^3/4

According to indices;

a^{m/n} = (\sqrt[n]{a}) ^m

Applying this to the question

64^{5/6} \\= (\sqrt[6]{64}) ^5\\= 2^5= 32

Similarly

27^{2/3}\\= (\sqrt[3]{27}) ^2\\= 3^2\\= 9\\

Also;

81^{3/4}\\= (\sqrt[4]{81}) ^3\\= 3^3\\= 27

Substitute inti the original equation

64^5/6 + 27^2/3 - 81^3/4

= 32 + 9- 27

= 41 - 27

= 14

Hence the required answer is 14

6 0
3 years ago
3(p+q)=p what’s answer
Blababa [14]

Answer:

Step-by-step explanation:

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7 0
4 years ago
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