Question

1. factor fully.

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

Explanation:

Step1: Use the Rational Root Theorem

The Rational Root Theorem states that if a polynomial \(P(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\) has a rational root \(p/q\), then \(p\) is a factor of \(a_0\) and \(q\) is a factor of \(a_n\).

Step2: Test possible roots for part a

For \(P(x)=x^{3}-6x^{2}+11x - 6\), \(a_n = 1\) and \(a_0=-6\). The possible rational roots are \(\pm1,\pm2,\pm3,\pm6\).
When \(x = 1\): \(P(1)=1^{3}-6\times1^{2}+11\times1 - 6=1 - 6+11 - 6=0\). So \((x - 1)\) is a factor.
Using long - division or synthetic division: \((x^{3}-6x^{2}+11x - 6)\div(x - 1)=x^{2}-5x + 6\).
Factor \(x^{2}-5x + 6=(x - 2)(x - 3)\). So \(P(x)=(x - 1)(x - 2)(x - 3)\).

Step3: Test possible roots for part b

For \(P(x)=x^{3}+2x^{2}-x - 2\), \(a_n = 1\) and \(a_0=-2\). The possible rational roots are \(\pm1,\pm2\).
When \(x = 1\): \(P(1)=1^{3}+2\times1^{2}-1 - 2=1 + 2-1 - 2=0\). So \((x - 1)\) is a factor.
Using synthetic division: \((x^{3}+2x^{2}-x - 2)\div(x - 1)=x^{2}+3x + 2\).
Factor \(x^{2}+3x + 2=(x + 1)(x + 2)\). So \(P(x)=(x - 1)(x + 1)(x + 2)\).

Step4: Test possible roots for part c

For \(P(v)=v^{3}+v^{2}-16v - 16\), \(a_n = 1\) and \(a_0=-16\). The possible rational roots are \(\pm1,\pm2,\pm4,\pm8,\pm16\).
When \(v=-1\): \(P(-1)=(-1)^{3}+(-1)^{2}-16\times(-1)-16=-1 + 1+16 - 16=0\). So \((v + 1)\) is a factor.
Using synthetic division: \((v^{3}+v^{2}-16v - 16)\div(v + 1)=v^{2}-16\).
Factor \(v^{2}-16=(v + 4)(v - 4)\). So \(P(v)=(v + 1)(v + 4)(v - 4)\).

Step5: Test possible roots for part d

For \(P(x)=x^{4}+4x^{3}-7x^{2}-34x - 24\), \(a_n = 1\) and \(a_0=-24\). The possible rational roots are \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24\).
When \(x=-1\): \(P(-1)=(-1)^{4}+4\times(-1)^{3}-7\times(-1)^{2}-34\times(-1)-24=1-4 - 7 + 34-24=0\). So \((x + 1)\) is a factor.
Using synthetic division: \((x^{4}+4x^{3}-7x^{2}-34x - 24)\div(x + 1)=x^{3}+3x^{2}-10x - 24\).
Testing possible roots for \(x^{3}+3x^{2}-10x - 24\), when \(x = 3\): \(3^{3}+3\times3^{2}-10\times3 - 24=27+27-30 - 24=0\).
So \((x - 3)\) is a factor of \(x^{3}+3x^{2}-10x - 24\). Using synthetic division again: \((x^{3}+3x^{2}-10x - 24)\div(x - 3)=x^{2}+6x + 8\).
Factor \(x^{2}+6x + 8=(x + 2)(x + 4)\). So \(P(x)=(x + 1)(x - 3)(x + 2)(x + 4)\).

Step6: Test possible roots for part e

For \(P(k)=k^{5}+3k^{4}-5k^{3}-15k^{2}+4k + 12\), \(a_n = 1\) and \(a_0=12\). The possible rational roots are \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\).
When \(k = 1\): \(P(1)=1^{5}+3\times1^{4}-5\times1^{3}-15\times1^{2}+4\times1+12=1 + 3-5 - 15+4 + 12=0\). So \((k - 1)\) is a factor.
Using synthetic division: \((k^{5}+3k^{4}-5k^{3}-15k^{2}+4k + 12)\div(k - 1)=k^{4}+4k^{3}-k^{2}-16k - 12\).
Testing possible roots for \(k^{4}+4k^{3}-k^{2}-16k - 12\), when \(k=-1\): \((-1)^{4}+4\times(-1)^{3}-(-1)^{2}-16\times(-1)-12=1-4 - 1 + 16-12=0\).
So \((k + 1)\) is a factor of \(k^{4}+4k^{3}-k^{2}-16k - 12\). Using synthetic division: \((k^{4}+4k^{3}-k^{2}-16k - 12)\div(k + 1)=k^{3}+3k^{2}-4k - 12\).
Testing possible roots for \(k^{3}+3k^{2}-4k - 12\), when \(k = 2\): \(2^{3}+3\times2^{2}-4\times2-12=8 + 12-8 - 12=0\).
So \((k - 2)\) is a factor of \(k^{3}+3k^{2}-4k - 12\). Using synthetic division: \((k^{3}+3k^{2}-4k - 12)\div(k - 2)=k^{2}+5k + 6\).
Factor \(k^{2}+5k + 6=(k + 2)(k + 3)\). So \(P(k)=(k - 1)(k + 1)(k - 2)(k + 2)(k + 3)\).

Answer:

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