X^3 – 6x^2 + 2x – 4, G(x) = 1 – 2x - Sarthaks EConnect

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x^3 – 6x^2 + 2x – 4, g(x) = 1 – 2x ← Prev Question Next Question → 0 votes 24.7k views

asked Feb 13, 2020 in Polynomials by ShasiRaj (56.4k points) edited Feb 14, 2020 by ShasiRaj

Using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division :

x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

factorization of polynomials
class-9

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1 Answer

+1 vote

answered Feb 13, 2020 by Tahseen Ahmad (30.6k points) selected Feb 14, 2020 by ShasiRaj Best answer

f(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

Put g(x) = 0

⇒ 1 – 2x = 0 or x = \(\frac{1}{2}\)

Remainder = f(\(\frac{1}{2}\))

Now,

f(\(\frac{1}{2}\)) = (\(\frac{1}{2}\))3 – 6(\(\frac{1}{2}\))2 + 2(\(\frac{1}{2}\)) – 4

= 1 + \(\frac{1}{8}\) – 4 – \(\frac{3}{2}\)

= \(\frac{-35}{8}\)

Actual Division:

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