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Stella [2.4K]
3 years ago
7

Find the remainder in the following division problem: x^5-4x^4+x^3-7x+1/x+2

Mathematics
1 answer:
icang [17]3 years ago
4 0

P(x)=x^5-4x^4+x^3-7x+1,\ Q(x)=x+2\\\\P(x)=H(x)\cdot Q(x)+R

Since the polynomial degree Q(x) is 1, the remainder of the division must be a number. Therefore we only need to calculate the value of polynomial for x = -2:

P(-2)=(-2)^5-4(-2)^4+(-2)^3-7(-2)+1\\\\=-32-4(16)+(-8)+14+1\\\\=-32-64-8+15\\\\=-89

Answer: The ramainder is equal -89.

x^5-4x^4+x^3-7x+1=H(x)(x+2)-89

Calculated in the program:

x^5-4x^4+x^3-7x+1=(x^4-6x^3+13x^2-26x+45)(x+2)-89\\\Downarrow\\\dfrac{x^5-4x^4+x^3-7x+1}{x+2}=x^4-6x^3+13x^2-26x+45\ \ \ r(-89)

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