Determine the reminder when f(x)= (x-1)^2 + x^5 + x^4 - x^2 + 2x is divided by g(x)=x^3+3x^2+3x+1 ?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the remainder when f(x) = (x-1)^2 + x^5 + x^4 - x^2 + 2x is divided by g(x)=x^3+3x^2+3x+1.

f(x) = (x-1)^2 + x^5 + x^4 - x^2 + 2x

=> 1 + x^5 + x^4

=> x^5 + x^4 + 1

g(x) = x^3 + 3x^2 + 3x + 1

We can use long division for this:

x^3 + 3x^2 + 3x + 1 | x^5 + x^4 + 1                       | x^2 - 2x+3

...................................x^5 + 3x^4 + 3x^3 + x^2

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...........................................-2x^4 - 3x^3 - x^2 + 1

...........................................-2x^4 - 6x^3 - 6x^2 - 2x

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......................................................3x^3 + 5x^2 + 2x + 1

......................................................3x^3 + 9x^2 + 9x + 3

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.................................................................-4x^2 - 7x - 2

The remainder when f(x) is divided by g(x) is -4x^2 - 7x - 2

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll expand the square from f(x):

f(x) = x^2 - 2x + 1 + x^5 + x^4 - x^2 + 2x

We'll eliminate like terms:

f(x) = x^5 + x^4 + 1

We'll recognize in expression of g(x) the binomial raised to cube:

g(x) = (x+1)^3

Since the divisor is of 3rd order, the reminder has to be of 2nd order.

r=ax^2 + bx + c

We 'll write the reminder theorem:

f=g*q + r, where q is the quotient of the division.

We notice that x=1 is a multiple root of the polynomial g.

f(-1)=g(-1)*q(-1) + r(-1)

By substituting the value x=-1 into all polynomials, we'll obtain:

f(-1)= (-1)^5 + (-1)^4 + 1

f(-1)=-1+1+1

f(-1)=1

g(-1)= (-1+1)^3=0

r(-1)=a-b+c

So, we'll have:

1=0*q(-1) + a-b+c

1=a-b+c

Because of the fact that x=-1 is a multiple root, it will cancel the first derivative of the expression: f=g*c + r

f' =(g*c)' + r'

5x^4 +4 x^3=3*(x+1)^2*q+(x+1)^3*q' +2*ax+ b

By substituting the value x=-1 into all polynomials, we'll obtain:

5-4=-2*a+b

1=-2*a+b

We'll calculate the first derivative of the expression

5x^4 +4 x^3=3*(x+1)^2*q+(x+1)^3*q' +2*ax+ b

(5x^4 +4 x^3)'=(3*(x+1)^2*q+(x+1)^3*q' +2*ax+ b)'

5*4* x^3 + 4*3* x^2=6*(x+1)*q+3*(x+1)^2*q'+3*(x+1)^2*q'+(x+1)^3*q"+2a

By substituting the value x=-1 into all polynomials, we'll obtain:

-20+12=2*a

We'll divide by 2:

a = -10 + 6

a = -4

-2a+b=1

-2*(-4) + b = 1

b = 1 - 8

b = -7

a-b+c = 1

-4+7+c=1

3+c=1

c=1-3

c=-2

The reminder is: r(x) = -4x^2 - 7x - 2.

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