For what value of k is (x + 1) is a factor of x 3+ kx 2– x + 2?
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Finding the Value of k in a Polynomial Factor Problem
The question asks us to find the value of a constant, k, in a cubic polynomial \(P(x) = x^3 + kx^2 – x + 2\), such that the linear expression \((x + 1)\) is a factor of this polynomial.
To solve this kind of problem, we can use the Factor Theorem. The Factor Theorem is a powerful tool in algebra that connects the roots of a polynomial to its factors.
Understanding the Factor Theorem
The Factor Theorem states that a polynomial \(P(x)\) has a factor \((x - a)\) if and only if \(P(a) = 0\). In simpler terms, if substituting a value 'a' into a polynomial makes the polynomial equal to zero, then \((x - a)\) is a factor of that polynomial.
Applying the Factor Theorem to Find k
In our specific problem, the given factor is \((x + 1)\). According to the Factor Theorem, if \((x + 1)\) is a factor of \(P(x) = x^3 + kx^2 – x + 2\), then substituting \(x = -1\) (because \(x+1 = x - (-1)\), so \(a = -1\)) into the polynomial \(P(x)\) must result in zero.
Let's substitute \(x = -1\) into the polynomial \(P(x)\):
\[P(-1) = (-1)^3 + k(-1)^2 – (-1) + 2\]
Now, we simplify this expression:
- \( (-1)^3 = -1 \)
- \( (-1)^2 = 1 \)
- \( -(-1) = +1 \)
Substituting these values back into the equation for \(P(-1)\):
\[P(-1) = -1 + k(1) + 1 + 2\]
\[P(-1) = -1 + k + 1 + 2\]
Combine the constant terms:
\[P(-1) = k + (-1 + 1 + 2)\]
\[P(-1) = k + 2\]
According to the Factor Theorem, since \((x + 1)\) is a factor, \(P(-1)\) must be equal to 0. So, we set the expression for \(P(-1)\) equal to 0:
\[k + 2 = 0\]
Now, we solve this simple linear equation for k:
Subtract 2 from both sides:
\[k = -2\]
Therefore, the value of k for which \((x + 1)\) is a factor of \(x^3 + kx^2 – x + 2\) is \(-2\).
Summary of Steps to Find k
- Identify the polynomial \(P(x) = x^3 + kx^2 – x + 2\) and the given factor \((x + 1)\).
- Use the Factor Theorem: If \((x+1)\) is a factor, then \(P(-1) = 0\).
- Substitute \(x = -1\) into \(P(x)\).
- Simplify the expression for \(P(-1)\).
- Set \(P(-1)\) equal to 0 and solve for k.
| Step | Calculation | Result |
|---|---|---|
| Substitute \(x=-1\) | \(P(-1) = (-1)^3 + k(-1)^2 - (-1) + 2\) | \(P(-1) = -1 + k + 1 + 2\) |
| Simplify | \(P(-1) = k + (-1 + 1 + 2)\) | \(P(-1) = k + 2\) |
| Set \(P(-1) = 0\) | \(k + 2 = 0\) | \(k = -2\) |
Polynomial Factor Problem Revision
This problem demonstrates a direct application of the Factor Theorem. When a linear expression \((x - a)\) is a factor of a polynomial \(P(x)\), it means that 'a' is a root of the polynomial, i.e., \(P(a) = 0\). Conversely, if \(P(a) = 0\), then \((x - a)\) is a factor of \(P(x)\).
For a factor \((x + 1)\), the corresponding value 'a' is \(-1\), because \((x + 1) = (x - (-1))\). Thus, to check if \((x + 1)\) is a factor or to make it a factor by finding a variable like k, we evaluate the polynomial at \(x = -1\) and set the result to zero.
Additional Information on Polynomials and Factors
Related to the Factor Theorem is the Remainder Theorem. The Remainder Theorem states that when a polynomial \(P(x)\) is divided by \((x - a)\), the remainder is \(P(a)\). The Factor Theorem is actually a special case of the Remainder Theorem. If the remainder \(P(a)\) is 0, it means \((x - a)\) divides \(P(x)\) evenly, which is the definition of \((x - a)\) being a factor.
Polynomial long division can also be used to solve this type of problem. If \((x + 1)\) is a factor of \(x^3 + kx^2 – x + 2\), then dividing \(x^3 + kx^2 – x + 2\) by \((x + 1)\) should yield a remainder of zero. The value of k could also be found by performing the division and setting the remainder expression to zero.
Understanding factors and roots is fundamental to solving polynomial equations and sketching polynomial graphs.
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