Divide a polynomial A?(x) = x4 – 7.5×3 + 14.5×2 + 3x – 20 by the monomial factor x – 2. Is x = 2 a root?
The Correct Answer and Explanation is:
To divide the polynomial A(x)=x4–7.5×3+14.5×2+3x–20A(x) = x^4 – 7.5x^3 + 14.5x^2 + 3x – 20
by the monomial factor x−2x – 2, we can use polynomial (or synthetic) division.
Step 1: Use Synthetic Division
We divide A(x)A(x) by x−2x – 2, so we use 2 in synthetic division.
Write down the coefficients of the polynomial: 1−7.514.53−201 \quad -7.5 \quad 14.5 \quad 3 \quad -20
Now perform synthetic division:
2 | 1 -7.5 14.5 3 -20
| 2 -11 7 20
----------------------------
1 -5.5 3.5 10 0
Result:
- Quotient:
x3−5.5×2+3.5x+10x^3 – 5.5x^2 + 3.5x + 10
- Remainder:
00
Since the remainder is zero, this means that x−2x – 2 is a factor, and therefore:
✅ x = 2 is a root of the polynomial A(x).
Explanation:
To determine whether x=2x = 2 is a root of the polynomial A(x)=x4−7.5×3+14.5×2+3x−20A(x) = x^4 – 7.5x^3 + 14.5x^2 + 3x – 20, we divide the polynomial by x−2x – 2. A root of a polynomial is a value of xx for which the polynomial evaluates to zero, meaning A(x)=0A(x) = 0.
A practical method for this is synthetic division, which is faster than long division when dividing by a linear binomial like x−cx – c. In this case, we use c=2c = 2 and write the coefficients of the polynomial: 1, -7.5, 14.5, 3, and -20.
Applying synthetic division, we compute a new row of values by multiplying the divisor (2) successively and adding to the next coefficient. The process ends with a final value, called the remainder. If the remainder is zero, then x−2x – 2 divides evenly into the polynomial, and x=2x = 2 is a root.
In this problem, the synthetic division yields a remainder of 0, which confirms that x−2x – 2 is a factor of A(x)A(x). Therefore, x=2x = 2 is a root of the polynomial.
The quotient from the division, x3−5.5×2+3.5x+10x^3 – 5.5x^2 + 3.5x + 10, is the remaining polynomial when x−2x – 2 is factored out. This result is useful for further factorization or solving higher-degree polynomials.
Thus, we conclude that x=2x = 2 is a valid solution to the equation A(x)=0A(x) = 0.
