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3 years ago

Determine if x - 2 is a factor of p(x) = x4 - 3x2 + 2x - 8, and explain why.

Mathematics

1 answer:

Natali [406]3 years ago

7 0

Answer:

If x-2 is a factor of p(x) = x^{4} - 3 x^{2} + 2 x - 8 ,then p(x) = 0

Solution:

According to factor theorem of polynomials,

x-a is a factor of polynomial p(x), if and only p(a)=0

Step 1:

Since x-2 is factor of p(x),

x - 2 = 0

x = 2

Step 2:

$p(x) = x^{4} - 3 x^{2} + 2 x - 8 ------ (equation 1)$

By substituting x = 2 in equation 1,

$p(x) = 2^{4} - 3(2)^{2} + 2(2) - 8$

p(x) = 16 - 3(4 ) + 4 -8

p(x) = 16 - 12 - 4

p(x) = 16 - 16

p(x) = 0

Since x - 2 is a factor of p(x) , we get p(x)=0

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A recursive rule for a sequence is given. Find the first four terms of the sequence.

tigry1 [53]

Answer:

So...

a0 = -2

a1 = (a0)2 - 4 = 4 - 4 = 0

a2 = (a1)2 - 4 = ((a0)2-4)2 - 4 =...= 02 - 4 = -4

For an+1, you may use the previous term (an) if you have just calculated it, rather than calculating it recursively again

a3 = (-4)2 - 4 = 12

The first four terms in the sequence are -2, 0, -4, 12.

Step-by-step explanation:

This recursive sequence is defined as follows:

a0 = -2 [the first term]

an+1 = an2 - 4 [for all other terms, n≥0; note that you cannot find a0 using this line]

Note that this is not "BASE" for a number base, but it is "SUB" for subscript, indicating the term. Sometimes, sequences are written with the first term being a1 and sometimes sequences are written with the first term being a0. This is because there are situations that make one or the other more convenient (for example, starting with elapsed time t=0 makes sense).

To find the value of later terms, a recursive definition requires that you find the value of the previous term, which requires that you find the value of the previous term, which requires that you find the value of the previous term, which requires that you find the value of the previous term, ... which requires that use the value of the first term.

That's why (joke) that the dictionary definition of "recursive" is "see recursive."

4 0

2 years ago

Ann runs a day care center. Of the last 18 children to enroll at the day care center, 8 of them have been elementary school stud

mrs_skeptik [129]

4 children i think. since you’re halving 18 you will half the 8 as well

4 0

3 years ago

Make q the subject of 4p=5q-2

kondor19780726 [428]

Answer:

5q = 4p + 2

Step-by-step explanation:

4p = 5q - 2

4p + 2 = 5q - 2 + 2

5q = 4p + 2

3 0

3 years ago

Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)

baherus [9]

Step 1

Given;

$4x^2-y^2+24x+4y+28=0$

Required; To find the center that eliminates the linear terms

Step 2

$\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}$

Step 3

Substitute a,d,e into the vertex form

$\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}$ $\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}$

Step 4

Completing the square for -y²+4y

$\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}$

Step 5

Substitute a,d,e into the vertex form

$\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}$

Step 6

$\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}$

Step 7

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}$

Hence the answer is (-3,2)

4 0

1 year ago

Use the graph of the polynomial function to find the factored form of the related polynomial. Assume it has no constant factor

Step2247 [10]

Answer:

Step-by-step explanation:

The factored from of a polynomial can be found from the zeros or x-intercepts of the graph.

The x-intercepts here are x= 1 and x= 7.

Then the factors are x-1 and x-7.

So the factored form is (x-1)(x-7).

6 0

3 years ago