Cting Polynomials: Practice

Mathematics

1 answer:

Colt1911 [192]2 years ago

7 0

The option that's true regarding the sum and difference of the polynomial p(x) and b is B. The sum and difference of p(x) and b are polynomials.

<h3>How to illustrate the information?</h3>

From the information given, it was stated that p(x) is a polynomial and is given as p(=) = a₁ + a₁ + a₂=² + ... + an.

In this case, it should be noted that the sum and difference of p(x) and b are polynomials.

They can't be integers as the function given is represented as a polynomial.

In conclusion, the correct option is B.

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I need to know the answer to p=4x-140

dalvyx [7]

This is not look like something that could be answered, check around the paper for a graphing box.
This looks like a formula for a line on a graph: y=Mx+b. I can help you if this is the case, 1. Put a point at -140 on the Y axis (up and down) 2. Move up 1 and over 4and put a dot there( you could multiply the 1 and 4 to cover a larger area) because it goes all the way to -140.

7 0

3 years ago

Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o

ivolga24 [154]

$\bf (2x+3)^5\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^5(+3)^0\\
2&+5&(2x)^4(+3)^1\\
3&+10&(2x)^3(+3)^2\\
4&+10&(2x)^2(+3)^3\\
5&+5&(2x)^1(+3)^4\\
6&+1&(2x)^0(+3)^5
\end{array}$

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element? well, 5*4/2
how did we get 10 for the fourth element? well, 10*2/4

$\bf (2x-3y)^4\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^4(-3y)^0\\
2&+4&(2x)^3(-3y)^1\\
3&+6&(2x)^2(-3y)^2\\
4&+4&(2x)^1(-3y)^3\\
5&+1&(2x)^0(-3y)^4
\end{array}$

$\bf (3a+4b)^8\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(3a)^8(+4b)^0\\
2&+8&(3a)^7(+4b)^1\\
3&+28&(3a)^6(+4b)^2\\
4&+56&(3a)^5(+4b)^3\\
5&+70&(3a)^4(+4b)^4\\
6&+56&(3a)^3(+4b)^5\\
7&+28&(3a)^2(+4b)^6\\
8&+8&(3a)^1(+4b)^7\\
9&+1&(3a)^0(+4b)^8
\end{array}$

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶

7 0

3 years ago

Read 2 more answers

4. Timothy deposited $3500 into an accountthat pays 4.3% simple interest and keeps it

hjlf

$3,500(4.3%)4= $602

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3 years ago

Note that 3 is a factor of 21. This means that 3×[]=21. Another way to write this is 3×n=21. Find a value of n that makes the st

mash [69]

Answer:

n=9