Write a polynomial function in standard form with zeros at 1, 2, and 3.

2 answers:

8 0

The polynomial whose zeroes are 1, 2 and 3 is given by,
(x-1)(x-2)(x-3) = 0
(x-1)[x²-3x-2x+6]=0
(x-1)[x²-5x+6]=0
x[x²-5x+6] -1[x²-5x+6]=0
x³-5x²+6x-x²+5x-6=0
x³-6x²+11x-6=0

Therefore, the required polynomial is,
x³ - 6x² + 11x -6 = 0

lara31 [8.8K]2 years ago

6 0

Answer: The required polynomial in standard form is $x^2-6x^2+11x-6.$

Step-by-step explanation: We are given to write a polynomial in standard form with zeroes 1, 2 and 3.

We know that

if a polynomial p(x) has zeroes at a, b and c, then its expression can be written as follows :

$p(x)=(x-a)(x-b)(x-c).$

Therefore, if f(x) represents the given polynomial, then we must have

$f(x)\\\\=(x-1)(x-2)(x-3)\\\\=(x^2-3x+2)(x-3)\\\\=x(x^2-3x+2)-3(x^2-3x+2)\\\\=x^3-3x^2+2x-3x^2+9x-6\\\\=x^2-6x^2+11x-6.$

Thus, the required polynomial in standard form is $x^2-6x^2+11x-6.$

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Describe and correct the error in solving the following rational equation , plz help!!!!!

aleksley [76]

First of all, x cannot ever equal -1 because if it does then in the original equation you'd be dividing by 0, which is not possible. anything divided by 0 is undefined.

I think the error was in the second step: x(x+1) = 2(x+1) because they didn't need to go from the first step to there. I'm assuming they multiplied both sides of the equation with (x+1) to get rid of the fraction, but since both sides have (x+1) as the denominator the (x+1) would've cancelled out instead of multiplying with the numerator.

Correction:

$\frac{x}{x+1} = \frac{2}{x+1}\\(\frac{x+1}{1}) (\frac{x}{x+1}) = (\frac{2}{x+1})(\frac{x+1}{1}) \\x = 2$