What must be added to x³-3x²-12x+19 so that the result is exactly divisible by x²+x-6 ?
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Answered by
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Given f(x) = x^3 - 3x^2 - 12x + 19.
Given g(x) = x^2 + x - 6.
Divide f(x) by g(x).
x^2 + x - 6) x^3 - 3x^2 - 12x + 19
x^3 + x^2 - 6x
-----------------------------
-4x^2 - 6x + 19
-4x^2 - 4x + 24
-----------------------------
-2x - 5
Here, remainder = -2x - 5.
Therefore (-2x - 5) must be added to x^3 - 3x^2 - 12x + 19, so that the result is divisible by x^2 + x - 6.
Hope this helps!
Given g(x) = x^2 + x - 6.
Divide f(x) by g(x).
x^2 + x - 6) x^3 - 3x^2 - 12x + 19
x^3 + x^2 - 6x
-----------------------------
-4x^2 - 6x + 19
-4x^2 - 4x + 24
-----------------------------
-2x - 5
Here, remainder = -2x - 5.
Therefore (-2x - 5) must be added to x^3 - 3x^2 - 12x + 19, so that the result is divisible by x^2 + x - 6.
Hope this helps!
:-)
Answered by
1
Given f(x) = x^3 - 3x^2 - 12x + 19.
Given g(x) = x^2 + x - 6.
Divide f(x) by g(x).
x^2 + x - 6) x^3 - 3x^2 - 12x + 19
x^3 + x^2 - 6x
-4x^2 - 6x + 19
-4x^2 - 4x + 24
-2x - 5
Here, remainder = -2x - 5.
Therefore (-2x - 5) must be added to x^3 - 3x^2 - 12x + 19, so that the result is divisible by x^2 + x - 6.
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