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3 years ago

How to Find a Greatest Common Factor in a Polynomial?

Mathematics

1 answer:

Lina20 [59]3 years ago

4 0

Answer:

To find the greatest common factor of a polynomial, first take all of the coefficients and list out their factors. After doing this, we can identify the greatest number that appears in each of the lists.

From there we will look for the lowest exponent for every variable. We'll use that number in our final answer. Remember, when doing this, that if the variable doesn't appear in a term, then the exponent is assumed to be 0.

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Please answer with picture thank you so much!

Ilia_Sergeevich [38]

Answer: 28

Step-by-step explanation: In this problem, we're given information about the parts of a segment and we're asked to find the value of x.

To find the value of x, we can use the <em>segment addition postulate</em>.

Looking at the diagram, it should make since that RS + ST = RT.

So 4x + 3x + 11 = 60 or 7x + 11 = 60.

Subtracting 11 from both sides and solving for x, we get x = 7.

Since RS is 4x, we can substitute a 7 in for x to get 4(7) or 28.

8 0

3 years ago

What is the solution to the system of equations graphed below? y = -2x - 3 y = 3x + 2 O A. (-1,1) B. (0,2)

Grace [21]

Answer:

A. (1-,1)

How ?: Just plug in -1 in the x spot and 1 in the y spot on both sides... If answers are the same that’s the answer.

3 0

3 years ago

Given that sec theta= -37/12, what is the value of cot theta , for pie/2 < theta< pie?

Triss [41]

The answer is B. Thank me in the future!

always remember ~Positive Vibes~

6 0

3 years ago

Read 2 more answers

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

7 0

3 years ago

Adiya said that the first step to solving the quadratic equation x2 + 6 = 20x by completing the square was to divide 6 by 2, squ

Anit [1.1K]

No, actually the method is not correct. Adiya needed to take x2 and subtract it from 20x. she would then get 6 = 18x. Have in mind also that she needed to divide 18. So 6 = 18x turns into 6=X. That is why is incorrect

7 0

3 years ago

Read 2 more answers