Chapter 4: Problem 72
Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=2 x^{3}-3 x^{2}-5 x+6 ; \quad k=1$$
Short Answer
Expert verified
The factorization of \(P(x)\) is \((x-1)(2x+3)(x-2)\).
Step by step solution
01
Use Synthetic Division
First, we will use the given zero \( k = 1 \) to perform synthetic division on \( P(x) = 2x^3 - 3x^2 - 5x + 6 \). Write down the coefficients [2, -3, -5, 6] and place \( k = 1 \) to the left in synthetic division. Proceed to carry down the 2, multiply and add through the row:\[\begin{array}{r|rrrr}1 & 2 & -3 & -5 & 6 \ & & 2 & -1 & -6 \\hline & 2 & -1 & -6 & 0 \\end{array}\]Since the remainder is 0, \( x - 1 \) is a factor of \( P(x) \), and the quotient is \( 2x^2 - x - 6 \).
02
Factor the Quotient Quadratic
Now, factor the quadratic quotient \( 2x^2 - x - 6 \). We need two numbers that multiply to \(-12\) (\(2 \times -6\)) and add to \(-1\). These numbers are \(-4\) and \(3\). Rewrite the middle term and factor by grouping:\[2x^2 - 4x + 3x - 6 = (2x^2 - 4x) + (3x - 6) = 2x(x - 2) + 3(x - 2) = (2x + 3)(x - 2)\]Thus, the quadratic \(2x^2 - x - 6\) factors into \((2x + 3)(x - 2)\).
03
Write the Complete Factorization
Since we've found that \( x - 1 \) is a factor, and the quadratic \(2x^2 - x - 6\) factors into \( (2x + 3)(x - 2) \), the complete factorization of \( P(x) \) is:\[ P(x) = (x - 1)(2x + 3)(x - 2) \].
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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 to divide a polynomial by a linear factor of the form \(x - k\). This technique significantly reduces the complexity and amount of work involved in polynomial division. For our exercise, we needed to factor the cubic polynomial \(P(x) = 2x^3 - 3x^2 - 5x + 6\), knowing that \(k = 1\) is a zero of the polynomial.To use synthetic division, you follow these steps:
- Write down the coefficients of the polynomial, in this case, [2, -3, -5, 6].
- Place the zero \(k\) to the left, in this case, 1.
- Carry down the leading coefficient (2) to the bottom row.
- Multiply the current number in the bottom row by \(k\), add this to the next coefficient above, and write the result in the bottom row.
- Repeat this process for each coefficient.
Factorization
Factorization involves breaking down a complex expression into simpler "factors" that can be multiplied to obtain the original expression. In polynomial terms, this means expressing the polynomial as a product of its polynomial factors. For the given polynomial \(P(x) = 2x^3 - 3x^2 - 5x + 6\), after using synthetic division, we ended up with the quotient polynomial \(2x^2 - x - 6\).To factor this quadratic polynomial, we need:
- Two numbers whose product equals the product of the leading coefficient (2) and the constant term (-6), which is -12.
- The same two numbers must add up to the coefficient of the \(x\) term, which is -1 in this case.
- Split the middle term: \(2x^2 - 4x + 3x - 6\).
- Group and factor: \(2x(x - 2) + 3(x - 2)\).
- Factor out the common term \((x - 2)\): \((2x + 3)(x - 2)\).
Zero of a Polynomial
The zero of a polynomial is a value of \(x\) that makes the polynomial equal to zero. In mathematical terms, if \(P(x)\) is a polynomial, then a number \(k\) is a zero if \(P(k) = 0\). This is critical because the zeros of a polynomial give you information about its factors and roots.For our problem, we began with the knowledge that \(k = 1\) is a zero of \(P(x) = 2x^3 - 3x^2 - 5x + 6\). This implies that \(x - 1\) is a factor of \(P(x)\) because substituting \(x = 1\) in the polynomial results in zero.Discovering a zero is paramount because:
- It helps to reduce the polynomial's degree when factored out.
- It gives insight into graphing the polynomial since these zeros are \(x\)-intercepts.
- Using the zero in synthetic division efficiently reveals remaining unfactored polynomial parts, leading to full factorization.