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polet [3.4K]
3 years ago
11

Is (x + 3) a factor of f(x) = 4x3 + 11x2 − 75x + 18?

Mathematics
2 answers:
Taya2010 [7]3 years ago
7 0

Answer:

No; the remainder is 0 so (x + 3) is not a factor

Step-by-step explanation:

harina [27]3 years ago
5 0

Answer: B) No; the remainder is 234, so (x+3) is not a factor.

===============================================================

Explanation:

We'll use the remainder theorem. That theorem says if we divide p(x) over (x-k), then the remainder is p(k). A special case of this theorem says that if we get 0 as the remainder, then (x-k) is a factor of p(x).

We're dividing f(x) over (x+3) which means that k = -3. Think of x+3 as x-(-3) so you can match it up with the form x-k.

To find the remainder of f(x)/(x+3), we need to compute f(-3).

Plug x = -3 into the f(x) function to get...

f(x) = 4x^3 + 11x^2 - 75x + 18

f(-3) = 4(-3)^3 + 11(-3)^2 - 75(-3) + 18

f(-3) = 234

The remainder is 234, which is isn't zero, so (x+3) is not a factor of f(x).

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Solve for y when x=28<br> x+3y=28
I am Lyosha [343]
Solving for y means to get that variable on one side of the equation and everything else needs to be on the opposite side of y.

We also know x is equal to 28. So, we can replace x with 28.
Now the equation will look something like 28 + 3y = 28

Now let us solve it.

28 + 3y = 28

First, subtract 28 from each side of the equation.

(28 + 3y) - 28 = 28 - 28 =
3y = 0

Now divide each side by 3.

3y / 3 = 0 / 3

y = 0/3

If a fraction has a numerator of 0, that means it's 0.

So, y is equal to 0.
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4 years ago
A math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of eac
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Answer:

<em><u>5</u></em><em><u> </u></em><em><u>points</u></em><em><u> </u></em><em><u>question</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>8</u></em>

<em><u>and</u></em><em><u> </u></em><em><u>2</u></em><em><u> </u></em><em><u>po</u></em><em><u>ints</u></em><em><u> questions</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>!</u></em><em><u>!</u></em>

Step-by-step explanation:

Let the number of 5 points questions be x and number of the 2 points questions be y !!

Given :- The paper has 38 problems

i.e,

number of 5 points Q + number of 2point Q = 38

=> x + y = 38

=> x = 38 - y .... ( i )

And , The test worth 100 pts :-

There must be x number of Questions which carry 5 points and there must y number of questions which carries 2 points ( as assume above )

so, 5x + 2y = 100 .... ( ii )

Putting value of the x in ( ii )

5 ( 38 - y ) + 2y = 100

190 - 5y + 2y = 100

- 3y = - 90

y = 30

Putting value of y in ( i )

x = 38 - y

x = 38 - 30

x = 8

So, the Number of 5points Question is

x = 8

and , the number of 2 points questions is y = 30 !!

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