If the polynomial x⁴ – 6x³ – 16x² + 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
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Dear Student:
Given: polynomial x⁴ – 6x³ – 16x² + 25x + 10 is divided by another polynomial x² – 2x + k,
the remainder comes out to be x + a,
To find k and a.
See the attachment.
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Solution :
x²-2x+k)x⁴-6x³-16x²+25x+10(x²-4x-(24+k)
**********x⁴-2x³+kx²
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************-4x³-(16+k)x²+25x
*********** -4x³+8x² - 4kx
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*************-(24+k)x²+(25+4k)x+10
**********-(24+k)x² +2(24+k)x-k(24+k)
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*************(25+4k-48-2k)x+10+k(24+k)
Remainder = ( -23+2k)x + (k²+24k+10)
Remainder = ( x + a ) [ Given ]
-23 + 2k = 1
=> 2k = 1 + 23
=> 2k = 24
=> k = 12
and
k²+24k+10 = a
=> 12²+24×12+10 = a
=> 144+288+10 = a
=> 442 = a
Therefore ,
a = 442 ,
k = 12
•••••
x²-2x+k)x⁴-6x³-16x²+25x+10(x²-4x-(24+k)
**********x⁴-2x³+kx²
______________________
************-4x³-(16+k)x²+25x
*********** -4x³+8x² - 4kx
______________________
*************-(24+k)x²+(25+4k)x+10
**********-(24+k)x² +2(24+k)x-k(24+k)
___________________________
*************(25+4k-48-2k)x+10+k(24+k)
Remainder = ( -23+2k)x + (k²+24k+10)
Remainder = ( x + a ) [ Given ]
-23 + 2k = 1
=> 2k = 1 + 23
=> 2k = 24
=> k = 12
and
k²+24k+10 = a
=> 12²+24×12+10 = a
=> 144+288+10 = a
=> 442 = a
Therefore ,
a = 442 ,
k = 12
•••••
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