Chapter 3: Problem 79
Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=x^{3}-2 x^{2}-7 x-4 ; \quad k=-1$$
Short Answer
Expert verified
The linear factors of \( P(x) \) are \((x + 1)^2(x - 4)\).
Step by step solution
01
Confirm k as a Zero
To confirm that \( k = -1 \) is a zero of the polynomial \( P(x) = x^3 - 2x^2 - 7x - 4 \), substitute \( x = -1 \) into \( P(x) \) and check if the result is 0.\[ P(-1) = (-1)^3 - 2(-1)^2 - 7(-1) - 4 = -1 - 2 + 7 - 4 = 0. \]Since \( P(-1) = 0 \), \( k = -1 \) is indeed a zero of \( P(x) \).
02
Perform Synthetic Division
Use synthetic division to divide \( P(x) \) by \( (x + 1) \), since \( x = -1 \) is a root. Write down the coefficients of \( P(x) \): \[ 1, -2, -7, -4. \]Perform synthetic division:\[\begin{array}{ccc|c} & -1 & \ \hline 1 & -2 & -7 & -4 \ & & -1 & 3 & 4 \ \hline & 1 & -3 & -4 & 0 \ \end{array}\]The quotient is \( x^2 - 3x - 4 \) and there is no remainder.
03
Factor the Quotient
Factor the quadratic \( x^2 - 3x - 4 \). Look for two numbers that multiply to \(-4\) and add to \(-3\): they are \(-4\) and \(1\). Thus, \( x^2 - 3x - 4 \) can be factored as \((x - 4)(x + 1)\).
04
Combine All Linear Factors
The original polynomial \( P(x) = x^3 - 2x^2 - 7x - 4 \) can be expressed as the product of its linear factors, given that \( k = -1 \) is a zero and the quotient was factored. Therefore, the complete factorization is: \[ P(x) = (x + 1)(x - 4)(x + 1). \]
05
Simplify the Factors
Since \( (x + 1) \) appeared twice, we can express the polynomial more succinctly as: \[ P(x) = (x + 1)^2(x - 4). \]This is the final factorization of \( P(x) \) into its linear factors.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Synthetic Division
Synthetic division is a simplified method of dividing a polynomial by a linear factor of the form \(x - k\). It is a quicker and more efficient alternative to the long division method.
This method uses only the coefficients of the polynomial, making it less cumbersome. It's particularly useful when you know that a certain value is a zero of the polynomial.
Here's a simple way to think about synthetic division:
This method uses only the coefficients of the polynomial, making it less cumbersome. It's particularly useful when you know that a certain value is a zero of the polynomial.
Here's a simple way to think about synthetic division:
- Write down the coefficients of the polynomial in descending order of the powers.
- Next to the coefficients, write the zero, \(k\), of the divisor \((x - k)\).
- Bring down the first coefficient unchanged.
- Multiply this value by \(k\) and add it to the next coefficient.
- Repeat this process until you finish with all coefficients.
- The numbers you get at the bottom, except for the last one, represent the coefficients of the quotient polynomial. The last number is the remainder.
Zero of a Polynomial
A zero of a polynomial is a value that, when substituted for the variable, results in the zero value for the polynomial. In other words, if \(k\) is a zero of the polynomial \(P(x)\), then \(P(k) = 0\). Identifying zeros of polynomials is crucial because it leads to their factorization.
When you have a polynomial like \(P(x) = x^3 - 2x^2 - 7x - 4\), knowing that \(k = -1\) is a zero helps you to break down the polynomial into simpler components. You confirm this by substituting \(x = -1\) into the polynomial and checking if it equals zero, just as the exercise shows.
When you have a polynomial like \(P(x) = x^3 - 2x^2 - 7x - 4\), knowing that \(k = -1\) is a zero helps you to break down the polynomial into simpler components. You confirm this by substituting \(x = -1\) into the polynomial and checking if it equals zero, just as the exercise shows.
- If \(P(k) = 0\), \(x - k\) is a factor of the polynomial.
- Zeros reveal the roots of the polynomial equation \(P(x) = 0\).
- These roots are where the graph of the polynomial crosses or touches the x-axis.
Quadratic Factoring
Once a polynomial is reduced using synthetic division, it often results in a simpler quadratic. Factoring quadratics is generally more straightforward and is a critical step toward polynomial factorization.
Consider the quadratic \(x^2 - 3x - 4\). Factoring it involves finding two numbers that multiply to the constant term (-4) and add up to the linear coefficient (-3).
Here is a simple strategy to factor quadratics:
Quadratic factoring completes the polynomial factorization, helping originally complex problems transform into their simplest linear factors before solving.
Consider the quadratic \(x^2 - 3x - 4\). Factoring it involves finding two numbers that multiply to the constant term (-4) and add up to the linear coefficient (-3).
Here is a simple strategy to factor quadratics:
- List pairs of numbers that multiply to the constant term.
- Identify which pair adds up to the linear term.
- Rewrite the middle term using your two numbers.
- Factor by grouping.
Quadratic factoring completes the polynomial factorization, helping originally complex problems transform into their simplest linear factors before solving.