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If ax^3 bx^2 x - 6 has x 2 as a factor & leaves a remainder '4' when divided by (x - 2). Find the values of a & b.?
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If ax^3 bx^2 x - 6 has x 2 as a factor & leaves a remainder '4' ...
If x-2 is a factor, then we know that (x-2) multiplied by some other polynomial will give us the original polynomial. We can use long division or synthetic division to find this other polynomial:

2 │ a b 1 -6
-2a 2a-b b+2a+1 -2b-2a-6
-------------------
a-2b 3a-2b b+2a+1 -2b-2a-6

Therefore, the original polynomial can be expressed as:

ax^3 + bx^2 + x - 6 = (x-2)(a*x^2 + (3a-2b)*x + (b+2a+1)) + (-2b-2a-6)

Note that the remainder (-2b-2a-6) is also a polynomial, but of lower degree than the divisor (x-2).
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If ax^3 bx^2 x - 6 has x 2 as a factor & leaves a remainder '4' when divided by (x - 2). Find the values of a & b.?
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