Question

Did I get these right I'm unsure exactly what the question wants ​

Answer 1

I can't completely see the last one but it seems like they are all right the correct answers are below if not

f(1) = 0

f(-1)=5

f(-7)=0

Explanation:

As seen in the picture below we do the opposite sign of whatever the number in the binomial is when putting it in the equation

so in the first equation it will look like this

f(1) =

so now we replace every x with a 1

`3(1)^3+4(1)^2-5(1)-2`

3(1)+4(1)-5-2

3+4-5-2

7-7

0

so its

f(1) =0

Now we do the same for the next one

`2(-1)^4-(-1)^3+3(-1)^2-1`

2(1)-(-1)+3(1)-1

2+1+3-1

3+3-1

6-1

5

f(-1)=5

Now for the last one

`2(-7)^4+14(-7)^3-2(-7)^2-14(-7)`

2(2401)+14(-343)-2(49)+98

4802-4802-98+98

0+0

0

f(-7)=0

hope this helps

Answer 2

Yeah this is correct if you only need answers, not the process.

If you want to know how you got these answers, here is Explanation:

`\sf\\5.\ (3x^3+4x^2-5x-2)/(x-1)\\\textsf{Let f(x)}=3x^3+4x^2-5x-2\\\textsf{Comparing }x-1\textsf{ with }x-a,\textsf{ a = 1.}\\\textsf{Using the remainder theorem,}\\Remainder=f(a)=f(1)=3(1)^3+4(1)^2-5(1)-2=3+4-5-2=0`

`\sf\\6.\ (2x^4-x^3+3x^2-1)/(x+1)\\Solution:\\\textsf{Let f(x)}=2x^4-x^3+3x^2-1\\\textsf{Comparing }x-a\textsf{ with }x+1,\textsf{ a}=-1.\\\textsf{Using Remainder Theorem,}\\Remainder(R)=f(a)\\or,\ R=f(-1)\\or,\ R=2(-1)^4-(-1)^3+3(-1)^2-1=2+1+3-1=5`

`\sf\\7.\ (2x^4+14x^3-2x^2-14x)/(x+7)\\Solution:\\\textsf{Let f(x)}=2x^4+14x^3-2x^2-14x\\\textsf{Comparing }x+7\textsf{ with }x-a,\textsf{ a}=-7\\\textsf{Using Remainder Theorem,}\\R=f(a)\\or,\ R=f(-7)=2(-7)^4+14(-7)^3-2(-7)^2-14(7)=0`