Problem 32 - Algebra and Trigonometry

Question

Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=2 x^{3}+7 x^{2}+4 x-4 $$

Step-by-Step Solution

Answer

The rational zeros are \(-1, \frac{1}{2}, -4\). Polynomial: \((x+1)(2x-1)(x+4)\).

Step 1: List Possible Rational Zeros

According to the Rational Root Theorem, possible rational zeros of a polynomial are the ratios \(\frac{p}{q}\), where \(p\) is a factor of the constant term (-4) and \(q\) is a factor of the leading coefficient (2). The possible rational zeros are \(\pm 1, \pm 2, \pm 4, \pm \frac{1}{2}, \pm \frac{1}{4}\).

Step 2: Test Possible Rational Zeros

Using synthetic division, test each possible rational zero. Start with \(x = 1\) and proceed.ewline For \(x = 1\): Performing synthetic division does not give a remainder of 0.ewline For \(x = -1\): Performing synthetic division gives a remainder of 0.ewline Hence, \(x + 1\) is a factor.

Step 3: Perform Polynomial Division

Divide the original polynomial \(2x^3 + 7x^2 + 4x - 4\) by \(x + 1\) using synthetic division or long division. The quotient is \(2x^2 + 5x - 4\).

Step 4: Factor the Quadratic Quotient

Factor \(2x^2 + 5x - 4\) using either the quadratic formula or factoring methods.ewline The quadratic can be factored into \((2x - 1)(x + 4)\).

Step 5: Write the Polynomial in Factored Form

Using the results of our factorization, the original polynomial \(P(x)\) can be written as \((x + 1)(2x - 1)(x + 4)\).

Key Concepts

Rational Root Theorem

To find the rational zeros of a polynomial, we use the Rational Root Theorem. This theorem states that any possible rational root of a polynomial equation is represented as \( \frac{p}{q} \), where \( p \) and \( q \) are integers. Specifically, \( p \) is a factor of the constant term, while \( q \) is a factor of the leading coefficient. This approach helps in narrowing down the list of possible rational solutions, thus enabling more efficient testing of values.

• Constant term: -4
• Leading coefficient: 2

By applying the theorem to our polynomial, we determine the possible rational roots to be \( \pm 1, \pm 2, \pm 4, \pm \frac{1}{2}, \pm \frac{1}{4} \). These are derived by considering all combinations of factors of -4 divided by factors of 2.

synthetic division

Synthetic division is a simplified method of dividing polynomials, particularly useful when testing potential roots identified through the Rational Root Theorem. It only works for divisors of the form \( x - c \), where \( c \) is a suspected root. This efficient technique involves fewer steps than traditional long division and remains a key tool for evaluating polynomial expressions.With synthetic division, we tested each possible rational root against our polynomial \( P(x) = 2x^3 + 7x^2 + 4x - 4 \). When \( x = -1 \) was tested, it produced a remainder of zero, indicating that \( x + 1 \) is a factor of the polynomial. It simplifies the process by organizing the operation into a neat series of arithmetic steps, ensuring clear identification of actual factors.

polynomial division

Once a factor is determined, polynomial division comes into play to simplify the polynomial further. Whether using synthetic division or long division, this method allows us to break down the polynomial by dividing it by the binomial factor discovered earlier. For our polynomial, dividing \( 2x^3 + 7x^2 + 4x - 4 \) by \( x + 1 \) results in reducing it to a second-degree polynomial. Specifically, the quotient becomes \( 2x^2 + 5x - 4 \). This simplification is crucial, as it reduces the degree of the polynomial, making it easier to factor further and identify remaining zeros.

factoring quadratics

Once the polynomial has been reduced via division, the next step is to factor the quadratic quotient. This can often be achieved through several methods, including trial and error, completing the square, or using the quadratic formula. In our case, the quadratic function \( 2x^2 + 5x - 4 \) is factorable by inspection, leading to its factorization as \( (2x - 1)(x + 4) \).

• Check factors by expanding: \( (2x - 1)(x + 4) = 2x^2 + 5x - 4 \)
• Verify that no further factorization is possible

When compiled together with the linear factor found earlier, the original polynomial \( P(x) \) can finally be expressed in its completely factored form as \( (x + 1)(2x - 1)(x + 4) \). This representation reveals all its zeros at once, providing a full solution.