what must be subtracted from x^3-6x^2-15x+80 so that the result is exactly divisible by x^2 +x-12
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ironman777:
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divide X^3 - 6X^2 - 15X + 80 BY X^2 + X -12
AFTER DIVIDING THAT YOU GOT A REMANIDER WHICH IS 4X - 4.
THEN YOU HAVE TO SUBTRACT THE REMAINDER WHICH IS 4X - 4 FROM X^3 - 6X^2 - 15X + 80.
AFTER DOING THIS I THINK YOU GET YOUR ANSWER
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Answered by
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Answer:
Let ax + b be subtracted from p(x)=x
3
−6x
2
−15x+80
so that it is exactly divisible by x
2
+x−12.
∴s(x)=x
3
−6x
2
−15x+80−(ax+b)
=x
3
−6x
2
−(15+a)x+(80−b)
Dividend = Divisor × quotient + remainder
But remainder will be zero.
∴ Dividend = Divisor × quotient
⇒s(x)=(x
2
+x−12)×quotient
⇒s(x)=x
3
−6x
2
−(15+a)x+(80−b)
Hence, x(4−a)+(−4−b)=0,x
=0
⇒4−a=0 & (−4−b)=0
⇒a=4 and b=−4
Hence, if in p(x) we subtract 4x - 4 then it is exactly divisible by x
2
+x−12
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