Final answer:
The quotient of the polynomial (x^3 + 7x^2 + 3x - 6) divided by (x - 1) is x^2 + 6x - 3, which can be done using the reverse tabular method. Therefore, the original polynomial can be expressed as (x - 1)(x^2 + 6x - 3).
Step-by-step explanation:
The student has asked to find the quotient of the following polynomial function using the reverse tabular method:
(x^3 + 7x^2 + 3x - 6) / (x - 1)
Then, express the polynomial as a product of two factors.
Firstly, we set up the reverse tabular method with the divisor (x-1) to the left and the dividend (x^3 + 7x^2 + 3x - 6) atop the columns. We fill in the coefficients of the dividend in descending order of powers of x. Since we're dividing by (x - 1), we'll bring down the leading coefficient of the dividend as is, which is the first term of the quotient:
Next, we multiply this term by the divisor (x-1) and write the result underneath the next coefficient of the dividend. We then add the two numbers in that column to get the next coefficient of the quotient:
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- (1)*(x - 1) = x - 1
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- 7 + (-1) = 6
Repeat this process with the new coefficient:
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- (6)*(x - 1) = 6x - 6
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- 3 + (-6) = -3
And finally:
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- (-3)*(x - 1) = -3x + 3
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- -6 + 3 = -3
This gives us the complete quotient of the division:
x^2 + 6x - 3
Therefore, the original polynomial can be expressed as a product of two factors:
(x - 1)(x^2 + 6x - 3)
We've successfully used the reverse tabular method and found the quotient and the resulting factors of the given polynomial.