Problem 71 - A Graphical Approach to College Algebra

Question

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=2 x^{3}-3 x^{2}-17 x+30 ; \quad k=2$$

Step-by-Step Solution

Answer

\(P(x) = (x - 2)(2x - 5)(x + 3)\).

Step 1: Verify that k is a zero

Evaluate the polynomial \(P(x)\) at \(x = k\) to verify that \(k = 2\) is indeed a zero. Substitute \(x = 2\) into the polynomial: \[P(2) = 2(2)^3 - 3(2)^2 - 17(2) + 30 \] Simplify it step by step: \[= 2(8) - 3(4) - 34 + 30 \]\[= 16 - 12 - 34 + 30 \]\[= 0\] Since \(P(2) = 0\), \(k = 2\) is a zero of \(P(x)\).

Step 2: Synthetic Division

Use synthetic division to divide \(P(x)\) by \(x - 2\), since \(k = 2\) is a zero.The coefficients of \(P(x) = 2x^3 - 3x^2 - 17x + 30\) are: 2, -3, -17, 30.Set up the synthetic division with 2 outside:\[\begin{array}{r|rrrr}2 & 2 & -3 & -17 & 30 \ & & 4 & 2 & -30 \ \hline & 2 & 1 & -15 & 0 \\end{array}\]The bottom row gives the quotient: \(2x^2 + x - 15\) and remainder 0, confirming that \(x - 2\) divides \(P(x)\) exactly.

Step 3: Factor the Quadratic

Now factor the resulting quadratic \(2x^2 + x - 15\). We need two numbers that multiply to \(2 \times -15 = -30\) and add to \(1\) (the coefficient of \(x\)). These numbers are 6 and -5.Rewrite and factor by grouping:\[2x^2 + 6x - 5x - 15 \]Group the terms:\[(2x^2 + 6x) + (-5x - 15)\]Factor each group:\[2x(x + 3) - 5(x + 3)\]Factor out the common factor \((x + 3)\):\[(2x - 5)(x + 3)\]

Step 4: Combine Factors

Combine the linear factor found from synthetic division \((x - 2)\) with the factors of the quadratic \((2x - 5)(x + 3)\) to express \(P(x)\) fully in linear factors:\(P(x) = (x - 2)(2x - 5)(x + 3)\).

Key Concepts

Synthetic Division Simplified

Synthetic division is a quick method for dividing polynomials, especially when you're dividing by a linear factor like \(x - k\). It's simpler and faster than long division, making it a preferred technique in many scenarios.
Here’s how to apply synthetic division:

• Write down the coefficients of the polynomial you wish to divide, in this case, \(2x^3 - 3x^2 - 17x + 30\).
• Place the zero of the divisor, \(k = 2\), outside of the division symbol.
• Bring down the leading coefficient, which is \(2\) in this problem.
• Multiply \(2\) by \(k\), add it to the next coefficient to get a new row entry, and repeat this process.

For our example, synthetic division showed that \(x - 2\) is a divisor of \(P(x)\), yielding a quotient of \(2x^2 + x - 15\). This step confirmed \(x - 2\) evenly divides \(P(x)\) with a remainder of zero.

Understanding Zero of Polynomial

The zero of a polynomial refers to a solution of the equation \(P(x) = 0\). Finding this number, denoted as \(k\), is crucial when factoring polynomials.
To determine if \(k\) is a zero of \(P(x)\):

• Substitute \(k\) into the polynomial.
• Calculate \(P(k)\).

If \(P(k) = 0\), then \(k\) is indeed a zero. In our example, substituting \(k = 2\) into \(P(x)\) yields zero. This result confirms \(2\) is a zero of the polynomial \(P(x) = 2x^3 - 3x^2 - 17x + 30\). Knowing the zero allows us to perform synthetic division, simplifying the polynomial for further factorization.

Quadratic Factorization Unlocked

When you reduce a cubic polynomial via synthetic division, often you're left with a quadratic polynomial to factor. Here, after division, we have \(2x^2 + x - 15\). Factoring a quadratic requires finding two numbers that:

• Multiply to the product of the leading coefficient and the constant term (in this case \(2 \times -15 = -30\)).
• Add up to the middle term's coefficient (here, \(1\)).

For the quadratic \(2x^2 + x - 15\), those numbers are \(6\) and \(-5\). Rewrite the middle term using these numbers and factor by grouping:

• Group into \((2x^2 + 6x)\) and \((-5x - 15)\).
• Factor out common factors: \(2x(x + 3) - 5(x + 3)\).
• The common binomial \((x + 3)\) appears, allowing to factor into \((2x - 5)(x + 3)\).

Together with the linear factor obtained earlier, the fully factored form of the polynomial becomes \((x - 2)(2x - 5)(x + 3)\).