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WINSTONCH [101]
3 years ago
7

Use the Factor Theorem to determine whether the first polynomial is a factor of the second polynomial.

Mathematics
2 answers:
pashok25 [27]3 years ago
8 0

Answer:

The polynomial (x -5) is not a factor of second polynomial 3x^2 + 7x + 40

Step-by-step explanation:

Factor theorem states that if you divide a polynomial p(x) by a factor x -a of that polynomial, then you will get a zero remainder.

i.,e p(x) = (x-a)q(x)   which means that if x - a is a factor of p(x), then the remainder, when we do synthetic division by  x= a, will be zero.

Determine whether the first polynomial is a factor of the second polynomial.

Given the polynomial:  f(x)=3x^2 + 7x + 40

For  x-5 to be a factor of f(x)=3x^2 + 7x + 40, the factor theorems implies that x = 5 must be a zero of f(x).

Now, to test whether x-5  is a factor;

Set x -5 = 0

⇒x = 5

Then,

we will use synthetic division method to divide f(x) by x =5

you can see the figure as shown below in the attachment.

Since, the remainder is 150 which is not equal to zero, then Factor theorem says that (x-5) is not a  factor of 3x^2 + 7x + 40



zmey [24]3 years ago
3 0

Answer:

No, (x-5) is not a factor of 3x^2+7x+40

Explanation:

Factor theorem states that if (x-a) is a factor of the function f(x) then f(a) = 0.

We can use this theorem to check whether a polynomial is a factor of other polynomial or not.

Further Explanation:

Here, we have to check if (x-5) is a factor of 3x^2+7x+40 or not.

For this we can use the above mentioned factor theorem.

In our case,

a = 5

and f(x)=3x^2+7x+40

So, we find f(5) and see if it is zero or not. If f(5) = 0 then (x-5) must be the factor the polynomial.

f(5)=[tex]3(5)^2+7(5)+40\\\\=75+35+40\\\\=150\neq0

Since, f(5) is not zero. Hence, from factor theorem, (x-5) is not a factor of 3x^2+7x+40

Learn More:

brainly.com/question/12482195 (Answered by Kudzordzifrancis)

brainly.com/question/11378552 (Answered by Alinakincsem)

Keywords:

Factor theorem, Remainder theorem.

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Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}

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V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\

\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)

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