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Find the value of 
c so that the polynomial 
p(x) is divisible by 
(x+2).

p(x)=4x^(3)+cx^(2)+x+2

c=

Find the value of c c so that the polynomial p(x) p(x) is divisible by (x+2) (x+2) .\newlinep(x)=4x3+cx2+x+2 p(x)=4 x^{3}+c x^{2}+x+2 \newlinec= c=

Full solution

Q. Find the value of c c so that the polynomial p(x) p(x) is divisible by (x+2) (x+2) .\newlinep(x)=4x3+cx2+x+2 p(x)=4 x^{3}+c x^{2}+x+2 \newlinec= c=
  1. Using the Remainder Theorem: To determine the value of cc that makes the polynomial p(x)p(x) divisible by (x+2)(x+2), we can use polynomial long division or synthetic division. However, a simpler method is to use the Remainder Theorem, which states that if a polynomial f(x)f(x) is divisible by (xk)(x - k), then f(k)=0f(k) = 0. In this case, since we want p(x)p(x) to be divisible by (x+2)(x + 2), we set x=2x = -2 and solve for p(2)=0p(-2) = 0.
  2. Substituting x=2x = -2: Substitute x=2x = -2 into the polynomial p(x)=4x3+cx2+x+2p(x) = 4x^3 + cx^2 + x + 2.\newlinep(2)=4(2)3+c(2)2+(2)+2p(-2) = 4(-2)^3 + c(-2)^2 + (-2) + 2
  3. Calculating p(2-2): Calculate the value of p(2)p(-2).p(2)=4(8)+c(4)2+2p(-2) = 4(-8) + c(4) - 2 + 2p(2)=32+4c2+2p(-2) = -32 + 4c - 2 + 2p(2)=32+4cp(-2) = -32 + 4c
  4. Setting p(2)=0p(-2) = 0: Since we want p(x)p(x) to be divisible by (x+2)(x + 2), we set p(2)=0p(-2) = 0.\newline32+4c=0-32 + 4c = 0
  5. Solving for c: Solve for c.\newline4c=324c = 32\newlinec=324c = \frac{32}{4}\newlinec=8c = 8

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