Question

please help ………………. …………. ………… i already have the answer for part A but im having trouble with Parts B and C

Answer 1

In part B we must perform the following operation:

`(5a^3+4a^2-3a+2)+(a^3-3a^2+3a-9)`

The key here is to group the terms according to the power of a they have:

`(5a^3+4a^2-3a+2)+(a^3-3a^2+3a-9)=(5a^3+a^3)+(4a^2-3a^2)+(-3a+3a)+(2-9)`

Then, we can use the distributive property of the multiplication but in reverse:

`b\cdot a+c\cdot a=(b+c)\cdot a`

If we do this in each of the terms between parenthesis we get:

`\begin{gathered} (5a^3+a^3)+(4a^2-3a^2)+(-3a+3a)+(2-9)= \\ =(5+1)a^3+(4-3)a^2+(-3+3)a-7 \\ (5+1)a^3+(4-3)a^2+(-3+3)a-7=6a^3+a^2-7 \end{gathered}`

Then the answer for part B is:

`6a^3+a^2-7`

In part C we must simplify:

`(4y^3-2y+9)-(2y^3-3y^2+4y+7)`

Here is important to remember that a negative sign before a parenthesis means that you have to change the sign of all the terms inside it. Then we have:

`(4y^3-2y+9)-(2y^3-3y^2+4y+7)=4y^3-2y+9-2y^3+3y^2-4y-7`

Now we can do the same thing we did in part B, we group the terms according to the powers of y:

`4y^3-2y+9-2y^3+3y^2-4y-7=(4y^3-2y^3)+3y^2+(-2y-4y)+(9-7)`

Then we apply the distributive property in reverse:

`\begin{gathered} (4y^3-2y^3)+3y^2+(-2y-4y)+(9-7)=(4-2)y^3+3y^2+(-2-4)y+2 \\ (4-2)y^3+3y^2+(-2-4)y+2=2y^3+3y^2-6y+2 \end{gathered}`

Then the answer for part C is:

`2y^3+3y^2-6y+2`