A3x - x4 +a2x2 - ax3 factorise?

1 answer:

6 0

$a^3x-x^4+a^2x^2-ax^3=(a^3x+a^2x^2)-(ax^3+x^4)\\\\=a^2x(a+x)-x^3(a+x)=(a+x)(a^2x-x^3)\\\\=x(a+x)(a^2-x^2)=\boxed{x(a+x)(a-x)(a+x)=x(a-x)(a+x)^2}\\\\Used:\\\\\ a^2-b^2=(a-b)(a+b)$

You might be interested in

What two values of x are roots of the polynomial

mariarad [96]

So this is taking this equation and plugging it into the quadratic formula -b+- (sqrt (b^2)-(4)ac)/2a so your A value is 1 B value is 3 and C value is -5 so it’s B&E

5 0

3 years ago

Read my name to understand this?

Molodets [167]

Answer:

Hello :) 12345678910

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find a formula for the described function. A rectangle has perimeter 16 m. Express the area A of the rectangle as a function of

Alex17521 [72]

Part A:

Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.

Let the length of the other side be x, then

$2(L+x)=16 \\ \\ \Rightarrow L+x=8 \\ \\ \Rightarrow x=8-L$

Thus, if the length of one of the side is x, the length of the other side is 8 - L.

Hence, the area of the rectangle in terms of L is given by

$L(8 - L) = 8L-L^2$

Part B:

To find the domain of A

Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.

Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.

Hence the domain of the area are the set of values between 0 and 8 not inclusive.

Therefore,

$dom(Area)=0\ \textless \ L\ \textless \ 8 \ or \ (0, 8)$