(x^2-10x+30)/(x-5) all it says is dividing polynomials
1 answer:
At each step, you determine the quotient term by dividing the highest-degree term of the dividend by the highest degree term of the divisor. When you can't do that anymore, you're done and the remaining dividend is the remainder.
Here, the highest-degree dividend term is x^2 and the highest-degree divisor term is x. Their ratio, x^2/x = x, is the first quotient term. In the usual manner of long division, you subtract the product of this quotient term and the divisor to get the new dividend.
For this problem, the quotient is (x -5) and the remainder is 5. You can write the result as
(x² -10x +30)/(x -5) = (x -5) +5/(x -5)
Here, the highest-degree dividend term is x^2 and the highest-degree divisor term is x. Their ratio, x^2/x = x, is the first quotient term. In the usual manner of long division, you subtract the product of this quotient term and the divisor to get the new dividend.
For this problem, the quotient is (x -5) and the remainder is 5. You can write the result as
(x² -10x +30)/(x -5) = (x -5) +5/(x -5)
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Check the picture below, so let's check the equations below hmmm
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... L(x) = f'(1)(x -1) +f(1)
The derivative is ...
... f'(x) = 4x^3 +4x
so the slope of the tangent line is f'(1) = 4+4 = 8.
The value of the function at x=1 is
... f(1) = 1^4 +2·1^2 = 3
So, your linearization is ...
... L(x) = 8(x -1) +3
or
... L(x) = 8x -5
