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Remainder theorem If we divide the polynomials f (x) and G (x) by 2x ^ 2-3x-2, we can get the co expressions 2x + 3 and 4x-1 respectively, then the co expressions obtained by dividing f (x) - G (x) by 2x + 1 are
Let f (x) = (2x ^ 2-3x-2) * r (x) + 2x + 3G (x) = (2x ^ 2-3x-2) * q (x) + 4x-1 (where R (x), q (x) are positive coefficient polynomials) f (x) - G (x) = (2x ^ 2-3x-2) * (R (x) - Q (x)) + 2x + 3-4x + 1 = (2x + 1) (X-2) * (R (x) - Q (x)) - (2x + 1) + 5 = (2x + 1) ((X-2) * (R (x) - Q (x)) - 1) + 5 obviously
The topic of remainder theorem
f(x)=x^5+6x^4-4x^3+25x^2+30x+20
Then f (- 7) =?
The solution by remainder theorem
If f (x) is removed by X + 7, the result must be f (x) = g (x) (x + 7) + 6. At this time, f (- 7) = 6
The problem of remainder theorem
If we divide the polynomials f (x) and G (x) by 2x ^ 2-3x-2, we get the remainder formulas 2x + 3 and 4x-1 respectively, then the remainder formula of dividing f (x) - G (x) by 2x + 1 is?
f(x)=a(2x^2-3x-2)+2x+3
=a(2x=1)(x-2)+2x+3
g(x)=a(2x^2-3x-2)+4x-1
=a(2x=1)(x-2)+4x-1
A is divisible
therefore
Just calculate (2x + 3) / (2x + 1) = 1.2
(4x-1)/(2x+1)=2.-3
answer is 2-(-3)=5
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