Problem P(x)=x4−3x2+kx−2P(x)=x^4-3x^2+kx-2P(x)=x4−3x2+kx−2P, left parenthesis, x, right parenthesis, equals, x, start superscript, 4, end superscript, minus, 3, x, squared, plus, k, x, minus, 2 where kkkk is an unknown integer. P(x)P(x)P(x)P, left parenthesis, x, right parenthesis divided by (x−2)(x-2)(x−2)left parenthesis, x, minus, 2, right parenthesis has a remainder of 10101010. What is the value of kkkk? K=k=k=

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cristoshiwa cristoshiwa · Answer 1 · 2020-07-21T04:34:43+02:00

Answer: k = 4

Step-by-step explanation:

For this division, to determine the value of k, use the Remainder Theorem, which states that:

polynomial p(x) = dividend (x-a) * quotient Q(x) + remainder R(x)

Knowing the degree of quotient is

degree of Q = degree of p(x) - degree of (x-a)

For this case, Q(x) is a third degree polynomial.

Using the theorem:

[tex]x^{4}-3x^{2}+kx-2 = (x-2)(ax^{3}+bx^{2}+cx+d) + 10[/tex]

[tex]x^{4}-3x^{2}+kx-2 = ax^{4} + x^{3}(b-2a)+x^{2}(c-2a)+x(d-2c)-2d+10[/tex]

a = 1

b - 2a = 0 ⇒ b = 2

c - 2b = -3 ⇒ c = 1

-2d + 10 = -2 ⇒ d = 6

d - 2c = k ⇒ k = 4

Therefore, k = 4 and Q(x) = [tex]x^{4} -2x^{2} + 4x + 2[/tex]