Polynomial Topic Questions (1 Viewer)

[appleibeats]

Suppose that P(x) = x^3 + x^2 + 6x - 3

a) use the remainder theorem to find the remainder when P(x) is divided by x + 2i
b)Hence find the remainder when p(x) is divided by: i) x - 2i , ii) x^2 + 4

 

[KingOfActing]

Suppose that P(x) = x^3 + x^2 + 6x - 3

a) use the remainder theorem to find the remainder when P(x) is divided by x + 2i
b)Hence find the remainder when p(x) is divided by: i) x - 2i , ii) x^2 + 4

By the remainder theorem, the remainder when P(x) is divided by x - a is equal to P(a)

Hence:

a) Remainder = P(-2i) = -7 - 4i
b) i) Remainder = P(2i) = - 7 + 4i

ii) P(x) = (x^2 + 4)Q(x) + ax + b

P(-2i) = -2ia +b = -7 - 4i

Equate real and imaginary parts, we get that b = -7, a = 2

So the remainder is 2x - 7

 

[appleibeats]

The coefficients of the polynomials P(x) = ax^3 + bx + c are real and P(x) has a multiple zero at x = 1. When P(x) is divided by x + 1 the remainder is 4. Find the values of a, b, and c.

 

[parad0xica]

I'm guessing multiple zero means double zero in this question. i.e. P(1) = P'(1) = 0.

i.e. P(1) = a + b + c = 0 . . . (1)

i.e. P'(1) = 3a + b = 0 . . . (2)

By the Remainder Theorem, P(-1) = 4.

i.e. P(-1) = - a - b + c = 4 . . . (3)

Now we solve (1), (2) and (3) simultaneously.

(1) + (3) gives us c = 2.

Sub c = 2 into (3) and add it with (2) to attain a = 1.

Sub a = 1 and c = 2 into (1) gives birth to our lovely b = -3.

 

[19KANguy]

When the polynomial ax^4 + bx^3 + x^2 + 4x + 2 iis divided by 2x^2 + 2x - 1 the remainder is (2x + 3), find the values of a and b

 

[calamebe]

When the polynomial ax^4 + bx^3 + x^2 + 4x + 2 iis divided by 2x^2 + 2x - 1 the remainder is (2x + 3), find the values of a and b

 

[InteGrand]

When the polynomial ax^4 + bx^3 + x^2 + 4x + 2 iis divided by 2x^2 + 2x - 1 the remainder is (2x + 3), find the values of a and b

Another method.

 

[Mahan1]

When the polynomial ax^4 + bx^3 + x^2 + 4x + 2 iis divided by 2x^2 + 2x - 1 the remainder is (2x + 3), find the values of a and b

A different method is to play with sum and product of the roots.
let then from the info in the question we know
is divisible by
Note the roots of Q(x) are

let's called the other two
from sum of the roots we get



that means

finally the last relation between roots is :

That implies b= a and sub it into (3) we get a=b=2