find the remainder x³-6x²+2x-4/1-3x
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Answer:
Answer:-
p(x) =x {}^{3} - 6x + x - 4p(x)=x
3
−6x+x−4
p( \frac{1}{2} ) = ( \frac{1}{2} ) {}^{3} - 6( \frac{1}{2} ) {}^{2} + 2( \frac{1}{2 } ) - 4p(
2
1
)=(
2
1
)
3
−6(
2
1
)
2
+2(
2
1
)−4
= \frac{1}{8} - 6( \frac{1}{4} ) + 1 - 4=
8
1
−6(
4
1
)+1−4
= \frac{1}{8} \frac{ - 6}{8} + 1 - 4=
8
1
8
−6
+1−4
= \frac{1}{8} \frac{ - 3}{2} + 1 - 4=
8
1
2
−3
+1−4
Now take L.C.M. for both the denominators
= (1 - 12 + 8 - \frac{32}{8}) = \frac{ - 35}{8}=(1−12+8−
8
32
)=
8
−35
Remainder - 35/8
Now division - 6x² - 1
After division answer u get
Remainder is - 11x³ - 3
Checking the answer
- 11 \times 3 - 3−11×3−3
⇒ - 111/8 - 3⇒−111/8−3
⇒ (- 11 - \frac{24}{8} )⇒(−11−
8
24
)
⇒ \frac{ - 35}{8}⇒
8
−35
Hence Answer proved
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