If 5x 3+ 5x 2– 6x + 9 is divided by (x + 3), then the remainder is
-63
Understanding Polynomial Division and Remainders
We are asked to find the remainder when the polynomial \(P(x) = 5x^3 + 5x^2 - 6x + 9\) is divided by the linear expression \((x + 3)\). This type of problem can be efficiently solved using the Remainder Theorem.
Applying the Remainder Theorem for Division by (x + 3)
The Remainder Theorem is a key concept in polynomial algebra. It states that if a polynomial \(P(x)\) is divided by a linear polynomial \((x - a)\), the remainder is \(P(a)\).
In our problem, the divisor is \((x + 3)\). To use the Remainder Theorem, we need to write this in the form \((x - a)\). We can write \((x + 3)\) as \((x - (-3))\). Therefore, the value of \(a\) in this case is \(-3\).
According to the Remainder Theorem, the remainder when \(P(x)\) is divided by \((x + 3)\) is equal to \(P(-3)\). We need to substitute \(x = -3\) into the polynomial \(P(x)\) and evaluate the expression.
Calculating the Remainder using \(P(-3)\)
Let's substitute \(x = -3\) into the polynomial \(P(x) = 5x^3 + 5x^2 - 6x + 9\):
\[ P(-3) = 5(-3)^3 + 5(-3)^2 - 6(-3) + 9 \] First, we evaluate the powers of \(-3\):
- \((-3)^3 = (-3) \times (-3) \times (-3) = 9 \times (-3) = -27\)
- \((-3)^2 = (-3) \times (-3) = 9\)
Now, substitute these values back into the expression for \(P(-3)\):
\[ P(-3) = 5(-27) + 5(9) - 6(-3) + 9 \]
Perform the multiplications:
- \(5 \times (-27) = -135\)
- \(5 \times 9 = 45\)
- \(-6 \times (-3) = 18\)
Substitute these results back:
\[ P(-3) = -135 + 45 + 18 + 9 \]
Now, perform the additions and subtractions from left to right:
- \(-135 + 45 = -90\)
- \(-90 + 18 = -72\)
- \(-72 + 9 = -63\)
So, \(P(-3) = -63\).
According to the Remainder Theorem, the remainder when \(5x^3 + 5x^2 - 6x + 9\) is divided by \((x + 3)\) is \(-63\).
| Step | Calculation | Explanation |
|---|---|---|
| 1 | Identify \(P(x)\) and the divisor. | \(P(x) = 5x^3 + 5x^2 - 6x + 9\), Divisor = \((x + 3)\). |
| 2 | Find the value of 'a' from the divisor \((x-a)\). | \(x + 3 = x - (-3)\), so \(a = -3\). |
| 3 | Apply the Remainder Theorem. | Remainder = \(P(a) = P(-3)\). |
| 4 | Substitute \(x = -3\) into \(P(x)\). | \(P(-3) = 5(-3)^3 + 5(-3)^2 - 6(-3) + 9\). |
| 5 | Evaluate the expression. | \(P(-3) = 5(-27) + 5(9) - (-18) + 9 = -135 + 45 + 18 + 9 = -63\). |
| 6 | State the remainder. | The remainder is -63. |
Remainder Theorem Revision Table
| Concept | Description |
|---|---|
| Remainder Theorem | If a polynomial \(P(x)\) is divided by \((x-a)\), the remainder is \(P(a)\). |
| Factor Theorem | A special case of the Remainder Theorem: If \(P(a) = 0\), then \((x-a)\) is a factor of \(P(x)\). |
| Synthetic Division | A shorthand method for dividing polynomials by linear factors of the form \((x-a)\). The last number in the result is the remainder. |
Additional Information on Polynomial Division
While the Remainder Theorem is useful for finding just the remainder, polynomial long division or synthetic division can be used to find both the quotient and the remainder when dividing polynomials.
When dividing a polynomial \(P(x)\) by a non-zero polynomial \(D(x)\), there exist unique polynomials \(Q(x)\) (quotient) and \(R(x)\) (remainder) such that:
\[ P(x) = D(x) \cdot Q(x) + R(x) \]
where the degree of \(R(x)\) is less than the degree of \(D(x)\). In our case, since the divisor \((x+3)\) has degree 1, the remainder \(R(x)\) must have degree less than 1, meaning it's a constant.
The Remainder Theorem simplifies finding this constant remainder without performing the full division process when the divisor is linear.
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