Problem 75 - A Graphical Approach to Precalculus with Limits

Question

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=-6 x^{3}-13 x^{2}+14 x-3 ; \quad k=-3$$

Step-by-Step Solution

Answer

P(x) = (x + 3)(x - 1)(-6x - 1)

Step 1: Verify if k is a zero of P(x)

To check if \(k = -3\) is a zero of \(P(x)\), substitute \(x = -3\) into \(P(x)\) and check if the result equals zero:\[P(-3) = -6(-3)^3 - 13(-3)^2 + 14(-3) - 3.\]Calculated, this becomes:\[P(-3) = -6(-27) - 13(9) - 42 - 3 = 162 - 117 - 42 - 3 = 0.\]Since \(P(-3) = 0\), \(k = -3\) is indeed a zero of \(P(x)\).

Step 2: Perform Synthetic Division

Since \(k = -3\) is a zero, use synthetic division to divide \(P(x)\) by \(x + 3\):- Write the coefficients \([-6, -13, 14, -3]\).- Bring down \(-6\).- Multiply \(-3\) by \(-6\), add to \(-13\) to get \(5\).- Multiply \(-3\) by \(5\), add to \(14\) to get \(-1\).- Multiply \(-3\) by \(-1\), add to \(-3\) to get \(0\).The division gives \(-6x^2 + 5x - 1\) with a remainder of 0.

Step 3: Factor the Quotient Further

The result from synthetic division \(-6x^2 + 5x - 1\) can be factored:1. Look for two numbers that multiply to \(-6 \times (-1) = 6\) and add to \(5\). The numbers are \(6\) and \(-1\).2. Rewrite the middle term: \(-6x^2 + 6x - x - 1\).3. Factor by grouping: - Group \(-6x^2 + 6x\) as \(-6x(x - 1)\). - Group \(-x - 1\) as \(-1(x - 1)\).4. Factor out the common factor \((x - 1)\) to get \((x - 1)(-6x - 1)\).

Step 4: Combine Factors

Now that we have factored the quotient, combine it with the initial linear factor \((x + 3)\):\[P(x) = (x + 3)(x - 1)(-6x - 1)\].

Step 5: Write Final Answer

The polynomial \(P(x)\) is completely factored into linear factors. So, the answer is:\[P(x) = (x + 3)(x - 1)(-6x - 1)\].

Key Concepts

Synthetic Division

Synthetic Division is a simplified way of dividing a polynomial by a binomial of the form \(x - k\). It is quicker and involves fewer steps than the traditional long division method. Instead of dividing each term of the polynomial as with long division, we use only the coefficients of the polynomial. This method is particularly effective when you know one zero of the polynomial.
To use synthetic division, follow these steps:

• Write down the coefficients of the polynomial \(-6, -13, 14, -3\).
• Next, write the zero of the polynomial which, for this example, is \(k = -3\).
• Bring down the leading coefficient (\(-6\) in this case) as it is.
• Multiply this below value by \(k\), and add it to the next coefficient. Continue this process until you have gone through all coefficients.

This method reduces the polynomial while confirming our zero by resulting in a remainder of zero. For our equation, synthetic division yielded \(-6x^2 + 5x - 1\). We then obtained the factors \((x + 3)(-6x^2 + 5x - 1)\) easily.

Zero of a Polynomial

The zero of a polynomial is a value for \(x\) that makes the polynomial equal zero. In other words, if \(P(k) = 0\), then \(k\) is a zero of the polynomial \(P(x)\). Finding this value is crucial because it helps factor the polynomial. By knowing a zero, you can perform techniques like synthetic division to simplify the polynomial.
In our example, we are given that \(-3\) is a zero of \(P(x) = -6x^3 - 13x^2 + 14x - 3\). To verify this, which is step one in factoring a polynomial, substitute \(-3\) for \(x\) in the polynomial. If you perform the calculations correctly, you will find that the expression results in zero. Hence, \(x + 3\) becomes one of the factors of \(P(x)\). This step transforms complex polynomial division into more manageable units and aids in further factoring.

Factoring by Grouping

Factoring by Grouping is an approach where you rearrange parts of a polynomial and create groups that can be factored separately. This method is particularly handy when dealing with polynomials that have four terms, as it allows you to break them down into easier-to-handle parts.
To factor by grouping, in this case, \(-6x^2 + 5x - 1\), follow these steps:

• Look for two numbers that multiply to the product of the leading coefficient and the constant term \((-6)\times(-1)\) which is \(6\), and which also add to the middle coefficient (\(5\)). Here, the numbers are \(6\) and \(-1\).
• Rewrite the polynomial as \(-6x^2 + 6x - x - 1\).
• Group the first two terms and the last two terms separately: \(-6x(x - 1) - 1(x - 1)\).
• Factor out the common factor from each group: \((x - 1)(-6x - 1)\).

Thus, using factoring by grouping, we've turned \(-6x^2 + 5x - 1\) into \((x - 1)(-6x - 1)\). When combined with our earlier factor \((x + 3)\), this results in the completely factored polynomial \(P(x) = (x + 3)(x - 1)(-6x - 1)\). This method makes solving polynomials much clearer and more straightforward.