Ask question

Ask question

All categories

antoniya [11.8K]

2 years ago

13

According to the fundamental theorem of algebra which polynomial function has exactly 8 roots? A.F(x)=(3x^2-4x-5)(2x^6-5) B.f(x)

=(3x^4+2x)^4 C. (4x^2-7)^3 D. F(x)=(6x^8-4x^5-1)(3x^2-4)

1 answer:

choli [55]2 years ago

3 0

I think it’s a because I added from the therum

You might be interested in

Which of the following sentences has a relative clause?

Aloiza [94]

D. The audience applauded the ambassador, who spoke for one-half hour.

8 0

3 years ago

A 100 ft. ladder rests on top of a hook and ladder truck with its base 11 feet from the ground. When the angle of elevation of t

vichka [17]

100 plus 11 co sin 81

6 0

2 years ago

Round the number 296,386,482 to the nearest million

adelina 88 [10]

297,000,000 is your answer.

3 0

3 years ago

Read 2 more answers

What is the missing number

just olya [345]

8 is the missing number
Let’s say x is 2,
3(2)+8=14
8(2)-2=14

8 0

3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago