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Given a real zero of the polynomial, determine all other real zeros, and write the polynomial in terms of a product of linear and/or irreducible quadratic factors. Polynomial $$P(x)=x^{4}+4 x^{3}-2 x^{2}-12 x+9$$ Zero $$1 \text{(multiplicity 2 )}$$

Short Answer

Expert verified
The polynomial can be factored as \((x - 1)^2 (x + 3)^2\).

Step by step solution

01

Understand Given Zero and Its Multiplicity

We are given that \( x = 1 \) is a root with multiplicity 2 of the polynomial \( P(x) = x^4 + 4x^3 - 2x^2 - 12x + 9 \). This means \((x - 1)^2\) is a factor. Our task is to find the other factors and write the polynomial as a product of linear and/or irreducible quadratic factors.
02

Factor Out \((x - 1)^2\) from \( P(x) \)

Divide \( P(x) \) by \((x - 1)^2\) to find the quotient. This step involves polynomial long division. The dividend is \( x^4 + 4x^3 - 2x^2 - 12x + 9 \) and the divisor is \((x - 1)^2 = x^2 - 2x + 1\). Perform the division to get the quotient which will be the remaining factor.
03

Perform Polynomial Long Division

Divide the first term of the dividend, \( x^4 \), by the first term of the divisor, \( x^2 \), to get \( x^2 \). Multiply \((x^2 - 2x + 1)\) by \(x^2\) and subtract from the original polynomial:\[ x^4 + 4x^3 - 2x^2 - 12x + 9 - (x^4 - 2x^3 + x^2) = 6x^3 - 3x^2 - 12x + 9 \]Divide \( 6x^3 \) by \( x^2 \) to get \( 6x \). Multiply \( 6x \) by \(x^2 - 2x + 1\) and subtract:\[ 6x^3 - 3x^2 - 12x + 9 - (6x^3 - 12x^2 + 6x) = 9x^2 - 18x + 9 \]Finally, divide \( 9x^2 \) by \( x^2 \) to get \( 9 \). Multiply \( 9 \) by \(x^2 - 2x + 1\) and subtract:\[ 9x^2 - 18x + 9 - (9x^2 - 18x + 9) = 0 \]The polynomial long division shows that the quotient is \( x^2 + 6x + 9 \).
04

Factor the Remaining Quotient

The remaining factor \( x^2 + 6x + 9 \) can be factored as a perfect square trinomial. It factors into \((x + 3)^2\). Thus, the polynomial can be written as \((x - 1)^2 (x + 3)^2\).
05

Write Polynomial as a Product of Factors

Combine all factors into one expression: \( P(x) = (x - 1)^2 (x + 3)^2 \). This represents the polynomial as a product of linear factors, given all factors are real.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Long Division
Polynomial long division is a method used to divide a polynomial by another polynomial of lesser or equal degree. It's similar to the long division performed with numbers. The main objective is to determine the quotient and remainder when performing this division. Here, our goal is to divide
  • The dividend: the original polynomial, which in this case is \( x^4 + 4x^3 - 2x^2 - 12x + 9 \).
  • The divisor: derived from the root's multiplicity, given as \((x - 1)^2 = x^2 - 2x + 1\).
In the process, we:
  • Compare the highest degree terms of the dividend and the divisor.
  • Divide the leading term of the dividend by the leading term of the divisor to form part of the quotient.
  • Multiply the entire divisor by this quotient term and subtract from the original polynomial.
  • Repeat the process with the new polynomial (without the subtracted part) until reaching a remainder of zero or a degree lower than the divisor.
Successfully performing polynomial long division here reveals the quotient \( x^2 + 6x + 9 \). This quotient is a crucial step towards fully factoring the polynomial.
Multiplicity of Roots
Multiplicity of a root refers to the number of times a root, or solution, appears in a polynomial equation. If a polynomial has a root with a multiplicity greater than one, it means that the solution repeats that number of times. In our exercise, the given root \( x = 1 \) is noted to have a multiplicity of 2.

This implies several things:
  • The factor associated with \( x = 1 \) appears twice in the polynomial when expressed as product of its factors, hence \((x - 1)^2\).
  • Graphically, the curve of the polynomial touches the x-axis at this zero and turns back, indicating that it bounces off the axis at \( x = 1 \).
Understanding multiplicity is valuable because it helps us better describe the behavior of the polynomial function at its roots, as well as aids in the factorization process. It was foundational in determining one of the factors in the polynomial \( P(x) = (x - 1)^2 (x + 3)^2 \).
Perfect Square Trinomial
A perfect square trinomial is a specific type of polynomial that results from squaring a binomial. Recognizing and factoring these trinomials can simplify finding the roots of a polynomial. From the quotient of our long division, we have the trinomial expression \( x^2 + 6x + 9 \).

This expression fits the form of a perfect square trinomial because:
  • The first term \(x^2\) is \((x)^2\).
  • The last term \(9\) is \((3)^2\).
  • The middle term \(6x\) is two times the product of 3 and x, that is, \(2 \times 3 \times x\).
This trinomial can be rewritten as \((x + 3)^2\). Recognizing perfect square trinomials allows us to factor polynomials quickly and easily, simplifying our overall polynomial into a product of linear factors. In this context, it translates the remaining part of the polynomial into \((x + 3)^2\), leading to the full factorized form \((x - 1)^2 (x + 3)^2\).

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