polynomial is x^3 - x^2 -11x + 15
If 3 is the zero of the polynomial, when you divide this polynomial with x-3, the remainder has to be 0?
yes. If y=x^3 - x^2 -11x + 15, when y=0 the roots are on the axis.
To determine if 3 is a zero of the polynomial x^3 - x^2 -11x + 15, we need to check if the remainder obtained when dividing the polynomial by x-3 is 0.
To divide the polynomial by x-3, we can use polynomial long division. Here's how:
1. Write the polynomial in descending order: x^3 - x^2 -11x + 15.
2. Determine the divisor, which is x-3.
3. Divide the first term of the dividend (x^3) by the first term of the divisor (x) to get x^2, and write it as the first term of the quotient.
4. Multiply the divisor (x-3) by the quotient term (x^2), which gives x^3 - 3x^2.
5. Subtract the result obtained in step 4 from the dividend: (x^3 - x^2 -11x + 15) - (x^3 - 3x^2) = -2x^2 - 11x + 15.
6. Bring down the next term from the dividend, which is -2x^2.
7. Repeat steps 3-6 until you have divided all the terms.
8. The remainder is the last result obtained. In this case, it is -2x^2 - 11x + 15.
Now, to verify if 3 is a zero of the polynomial, we need to substitute x=3 into the remainder obtained in step 8 and check if the result is 0.
When x=3, the remainder becomes: -2(3)^2 - 11(3) + 15 = -2(9) - 33 + 15 = -18 - 33 + 15 = -36 - 33 = -69.
Since the remainder is -69 and not 0, 3 is not a zero of the polynomial x^3 - x^2 -11x + 15.
Therefore, the statement "If 3 is the zero of the polynomial, when you divide this polynomial with x-3, the remainder has to be 0" is not true in this case.