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makvit [3.9K]
4 years ago
10

Hey! i've been working on these questions but I have no idea how to solve this one, could anybody help me? Thanks in advance!

Mathematics
1 answer:
Aloiza [94]4 years ago
7 0

Answer:

1) \boxed{p(x) = x^3-x^2+x-1}

2) \boxed{p(x) = x^2+x-2}

3) \boxed{p(x) =- 2x^2+2x+4}

4) \boxed{p(x) = 2x^2+x-4}

Step-by-step explanation:

<u><em>Part (1)</em></u>

p(x) = x^3-x^2+x-1

As we have to determine it by ourselves, this is the polynomial having a degree of 3. p(x) with a degree of 3 means that the highest degree/exponent of x should be 3.

<u><em>Part (2)</em></u>

p(x) = x^2+x-2

This can be the polynomial having the factor x-1 because if we put:

x - 1 = 0 =>  x = 1 in the above polynomial, it gives us a result of zero which shows us that (x-1) "is" a factor of the polynomial.

<u><em>Part (3)</em></u>

p(x) = -2x^2+2x+4

This can be the polynomial for which p(0) = 4 and p(-1) = 0

<u>Let's check:</u>

p(0) =- 2(0)^2+2(0)+4\\p(0) = 0 + 0+4\\p(0) = 4

p(-1)= -2(-1)^2+2(-1)+4\\p(-1) = -2(1)-2+4\\p(-1) = -2-2+4\\p(-1) = 0

So, this is the required polynomial determined by "myself".

<u><em>Part (4):</em></u>

p(x) = 2x^2+x-4

This is the polynomial having a remainder 6 when divided by (x-2)

<u>Let's check:</u>

Let x - 2 = 0 => x = 2

<u>Putting in the above polynomial</u>

p(x) = 2(2)^2+(2)-4\\Given \ that \ Remainder = 6\\6 = 2(4) +2-4\\6 = 8+2-4\\6 = 10-4\\6 = 6

So, Proved that it has a remainder of 6 when divided by (x-2)

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