Question

Factor fully. a) \(P(x)=x^{3}-6 x^{2}+11 x-6\) b) \(P(x)=x^{3}+2 x^{2}-x-2\) c) \(P(v)=v^{3}+v^{2}-16 v-16\) d) \(P(x)=x^{4}+4 x^{3}-7 x^{2}-34 x-24\) e) \(P(k)=k^{5}+3 k^{4}-5 k^{3}-15 k^{2}+4 k+12\)

Short answer

a) \( (x-1)(x-2)(x-3) \) b) \( (x+1)(x-1)(x+2) \) c) \( (v+1)(v-4)(v+4) \) d) \( (x+2)(x+1)(x-3)(x+4) \) e) \( (k+2)(k+1)(k-1)(k-2)(k+3) \)

Step-by-step solution

Identify Possible Rational Roots

Use the Rational Root Theorem to list all potential rational roots by examining the factors of the constant term and the leading coefficient.

Test the Possible Roots

Substitute possible roots into the polynomial to check if they satisfy the equation, using synthetic division for efficiency.

Factor Using Found Roots

Once a root is found, use polynomial division or synthetic division to divide the original polynomial by the corresponding binomial. Repeat the process with the quotient polynomial until all factors are found.

Example a) Test Roots and Factor

Test possible roots for the polynomial, like \(x=1\). Since \(P(1)=0\), \(x-1\) is a factor. Now divide \(P(x)\) by \(x-1\) to get \((x-1)(x^2-5x+6)\). Further factor \(x^2-5x+6\) to get \((x-2)(x-3)\). The fully factored form is \(x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)\).

Example b) Test Roots and Factor

Test possible roots for the polynomial, like \(x=-1\). Since \(P(-1)=0\), \(x+1\) is a factor. Now divide \(P(x)\) by \(x+1\) to get \((x+1)(x^2+x-2)\). Further factor \(x^2+x-2\) to get \((x-1)(x+2)\). The fully factored form is \(x^3+2x^2-x-2 = (x+1)(x-1)(x+2)\).

Example c) Test Roots and Factor

Test possible roots for the polynomial, like \(v=-1\). Since \(P(-1)=0\), \(v+1\) is a factor. Now divide \(P(v)\) by \(v+1\) to get \( (v+1)(v^2-16) \). Further factor \(v^2-16\) as a difference of squares: \( (v+1)(v-4)(v+4) \). The fully factored form is \(v^3+v^2-16v-16 = (v+1)(v-4)(v+4)\).

Example d) Test Roots and Factor

Test possible roots for the polynomial, like \(x=-2\). Since \(P(-2)=0\), \(x+2\) is a factor. Now divide \(P(x)\) by \(x+2\) to get \((x+2)(x^3+2x^2-11x-12)\). Factor the cubic polynomial: test roots like \(x=-1\). Since \(P(-1)=0\), \(x+1\) is a factor. Now divide \(x^3+2x^2-11x-12\) by \(x+1\) to get \((x+2)(x+1)(x^2+x-12)\). Further factor \(x^2+x-12\) to get \((x-3)(x+4)\). The fully factored form is \(x^4+4x^3-7x^2-34x-24 = (x+2)(x+1)(x-3)(x+4)\).

Example e) Test Roots and Factor

Test possible roots for the polynomial, like \(k=-2\). Since \(P(-2)=0\), \(k+2\) is a factor. Now divide \(P(k)\) by \(k+2\) to get \((k+2)(k^4+k^3-7k^2-5k+6)\). Factor the quartic polynomial: test roots like \(k=-1\). Since \(P(-1)=0\), \(k+1\) is a factor. Now divide \(k^4+k^3-7k^2-5k+6\) by \(k+1\) to get \((k+2)(k+1)(k^3-8k+6)\). Further factor \(k^3-8k+6\): test roots like \(k=1\). Since \(P(1)=0\), \(k-1\) is a factor. Now divide \(k^3-8k+6\) by \(k-1\) to get \((k+2)(k+1)(k-1)(k^2+k-6)\). Further factor \(k^2+k-6\) to get \((k-2)(k+3)\). The fully factored form is \(k^5+3k^4-5k^3-15k^2+4k+12 = (k+2)(k+1)(k-1)(k-2)(k+3)\).

Key concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool for finding possible rational roots of a polynomial equation. The theorem states that any possible rational root of the polynomial \[a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\]must be a factor of the constant term \(a_0\) divided by a factor of the leading coefficient \(a_n\). This provides a finite list of potential candidates you can test by substitution. For instance, if we have a polynomial \(P(x) = x^3 - 6x^2 + 11x - 6\), possible rational roots are factors of \( 6 / 1 \) (i.e., ±1, ±2, ±3, ±6).Knowing this theorem helps narrow down the roots you need to check, making the process more straightforward.

Synthetic Division

Synthetic Division is a shortcut method for dividing polynomials when testing potential roots. It’s simpler and faster than long division. Here's how you do it:1. Write down the coefficients of the polynomial.2. Use the potential root you are testing (let's call it \(r\)).3. Perform the synthetic division steps: multiply, add, and repeat.For example, let's divide \(x^3 - 6x^2 + 11x - 6\) by \(x - 1\) (which tests if \( x = 1 \) is a root):Write down the coefficients: \[ 1, -6, 11, -6 \]Set up and perform synthetic division: the root: 1\[ 1 | 1 -6 11 -6 \]Multiply 1 by 1, add to -6, etc.Using synthetic division properly will reveal the quotient and any remainder, helping you factor effectively.

Polynomial Division

Polynomial Division, whether long division or synthetic division, is key to breaking down polynomials into their factors. The process involves dividing the polynomial by a binomial to see if it's a factor. If it divides evenly (remainder is zero), then the binomial is a factor.To divide \( P(x) = x^3 - 6x^2 + 11x - 6 \) by \(x - 1\), you:1. Set up the division as you would with numbers.2. Divide the leading term of \(P(x)\) by the leading term of \(x - 1\).3. Multiply the entire divisor by this quotient term and subtract.4. Repeat the process with the new polynomial.Found factors are used to further break down the polynomial until it's completely factored.

Factoring Cubic Polynomials

To factor cubic polynomials (polynomials of degree 3), you generally start by finding at least one root using the Rational Root Theorem, and then use polynomial or synthetic division. For instance, let's factor \(P(x) = x^3 - 6x^2 + 11x - 6\):1. Find the root, say \(x = 1\).2. Use synthetic division: dividing \(P(x)\) by \(x - 1\) gives us \((x - 1)(x^2 - 5x + 6)\).3. Further factor \(x^2 - 5x + 6\) to get \( (x - 2)(x - 3)\).So the fully factored form is: \( (x - 1)(x - 2)(x - 3) \).Understanding this process helps in breaking down complex polynomials into simpler, solvable parts.

Factoring Quartic Polynomials

Factoring Quartic Polynomials (polynomials of degree 4) can be more challenging but follows similar steps. You look for roots to simplify the polynomial step by step.For example, with \(P(x) = x^4 + 4x^3 - 7x^2 - 34x - 24\):1. Use the Rational Root Theorem to list possible rational roots.2. Test these roots using substitution or synthetic division, e.g., \(x = -2\).3. Perform synthetic division: dividing \(P(x)\) by \(x + 2\) leads us to \( (x + 2)(x^3 + 2x^2 - 11x - 12) \).4. Repeat the process for the cubic term: say \(x = -1\), leading to \( (x + 2)(x + 1)(x^2 + x - 12) \).5. Further factor \( x^2 + x - 12 \) to \( (x - 3)(x + 4) \).The fully factored form is \( (x + 2)(x + 1)(x - 3)(x + 4) \).Breaking down quartic polynomials requires patience, but following these steps systematically simplifies the process greatly.