Question

using remainder theorem find the reminder when x^3-6x^2+2x-4 divided by (1-2x)​

Answer 1

Step-by-step explanation:

let p(x)=x^3-6x^2+2x-4

first find the root of 1-2x

=>1-2x=0

=>x=1/2

now, by remainder theorem when p(x) is divided by 1-2x the remainder is

= p(1/2)=(1/2)³-6(1/2)² +2(1/2)-4

=1/8-3/2+1-4

=(1-12-24)/8

= -35/8ans

Answer 2

${\huge{\underline{\underline{\sf{\blue{SOLUTION:-}}}}}}$

${\huge{\underline{\underline{\sf{\red{Given:-}}}}}}$

⠀⠀⠀⠀• p(x) = x³-6x²+2x-4

⠀⠀⠀ • g(x) = 1-2x

${\huge{\underline{\underline{\sf{\red{To\:Calculate :-}}}}}}$

⠀⠀⠀⠀• Find Remainder __?

${\huge{\underline{\underline{\sf{\red{Explanation :-}}}}}}$

⠀⠀⠀⠀• p(x) = x³-6x²+2x-4

⠀⠀⠀ • g(x) = 1-2x

⠀⠀⠀⠀⠀➯ 1-2x=0

⠀⠀⠀⠀⠀➯ -2x=-1

⠀⠀⠀⠀⠀➯ x= -1/-2

⠀⠀⠀⠀⠀➯ x= 1/2

• p(x) = x³-6x²+2x-4

Putting x= 1/2 in p(x)

$☞ \: ( \frac{1}{2} ) {}^{2} - 6 \times ( \frac{1}{2} ) {}^{2} + 2 \times \frac{1}{2} - 4$

$☞ \: \frac{1}{8} - 6 \times \frac{1}{4} + 2 \times \frac{1 }{2} - 4$

$☞ \: \frac{1}{8} - \frac{3}{2} + \frac{1}{1} - 4$

Take LCM

$☞ \: \frac{1 - 12 + 8 - 32}{8}$

$☞ \: \frac{ - 44 + 9}{8}$

${\boxed{\red{\tt{☞ -35/8 }}}}$

${\huge{\underline{\underline{\sf{\blue{Hence :-}}}}}}$

⠀

⠀⠀⠀⠀ • Remainder is {-35/8}

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