Question

1. $f(x) = \frac{x^2 + x - 42}{x^5 - 3x^4 - 45x^3 + 27x^2 + 648x + 972}$

Explanation:

Step1: Factor Numerator

Factor \(x^2 + x - 42\). We need two numbers that multiply to \(-42\) and add to \(1\). These numbers are \(7\) and \(-6\). So, \(x^2 + x - 42=(x + 7)(x - 6)\).

Step2: Factor Denominator

Factor \(x^5-3x^4 - 45x^3+27x^2+648x + 972\). First, try rational roots using Rational Root Theorem. Possible roots are \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm27,\pm36,\pm54,\pm81,\pm108,\pm162,\pm324,\pm486,\pm972\).

Test \(x=-3\): \((-3)^5-3(-3)^4 - 45(-3)^3+27(-3)^2+648(-3)+972=-243 - 243+1215 + 243-1944 + 972=(-243-243-1944)+(1215 + 243+972)=(-2430)+(2430)=0\). So, \(x + 3\) is a factor.

Divide the polynomial by \(x + 3\) using polynomial long division or synthetic division.

Using synthetic division for \(x=-3\):

Coefficients: \(1\) (x^5), \(-3\) (x^4), \(-45\) (x^3), \(27\) (x^2), \(648\) (x), \(972\) (constant)

Bring down \(1\). Multiply by \(-3\): \(-3\). Add to next coefficient: \(-3+(-3)=-6\). Multiply by \(-3\): \(18\). Add to next coefficient: \(-45 + 18=-27\). Multiply by \(-3\): \(81\). Add to next coefficient: \(27+81 = 108\). Multiply by \(-3\): \(-324\). Add to next coefficient: \(648+(-324)=324\). Multiply by \(-3\): \(-972\). Add to last coefficient: \(972+(-972)=0\).

So, the quotient is \(x^4-6x^3-27x^2+108x + 324\).

Now, factor \(x^4-6x^3-27x^2+108x + 324\). Try \(x=-3\) again: \((-3)^4-6(-3)^3-27(-3)^2+108(-3)+324=81 + 162-243-324 + 324=(81 + 162)+(-243-324 + 324)=243-243 = 0\). So, \(x + 3\) is a factor again.

Divide \(x^4-6x^3-27x^2+108x + 324\) by \(x + 3\) using synthetic division for \(x=-3\):

Coefficients: \(1\) (x^4), \(-6\) (x^3), \(-27\) (x^2), \(108\) (x), \(324\) (constant)

Bring down \(1\). Multiply by \(-3\): \(-3\). Add to next coefficient: \(-6+(-3)=-9\). Multiply by \(-3\): \(27\). Add to next coefficient: \(-27 + 27=0\). Multiply by \(-3\): \(0\). Add to next coefficient: \(108+0 = 108\). Multiply by \(-3\): \(-324\). Add to last coefficient: \(324+(-324)=0\).

Quotient: \(x^3-9x^2+0x + 108=x^3-9x^2+108\). Wait, no, wait, the coefficients after division: \(x^3-9x^2+0x + 108\)? Wait, no, the synthetic division steps:

After first step (bring down 1), multiply by -3: -3. Add to -6: -9. Multiply by -3: 27. Add to -27: 0. Multiply by -3: 0. Add to 108: 108. Multiply by -3: -324. Add to 324: 0. So the quotient is \(x^3-9x^2+0x + 108=x^3-9x^2+108\)? Wait, no, that can't be. Wait, the original polynomial after first division was \(x^4-6x^3-27x^2+108x + 324\). Dividing by \(x + 3\) gives \(x^3-9x^2+0x + 108\)? Wait, let's check by multiplying \((x + 3)(x^3-9x^2+0x + 108)=x^4-9x^3+0x^2+108x+3x^3-27x^2+0x + 324=x^4-6x^3-27x^2+108x + 324\). Yes, correct.

Now, factor \(x^3-9x^2+108\). Try \(x = 6\): \(6^3-9(6)^2+108=216-324 + 108=0\). So, \(x - 6\) is a factor.

Divide \(x^3-9x^2+108\) by \(x - 6\) using synthetic division for \(x = 6\):

Coefficients: \(1\) (x^3), \(-9\) (x^2), \(0\) (x), \(108\) (constant)

Bring down \(1\). Multiply by \(6\): \(6\). Add to \(-9\): \(-3\). Multiply by \(6\): \(-18\). Add to \(0\): \(-18\). Multiply by \(6\): \(-108\). Add to \(108\): \(0\). So the quotient is \(x^2-3x - 18\).

Factor \(x^2-3x - 18\): numbers that multiply to \(-18\) and add to \(-3\) are \(-6\) and \(3\). So, \(x^2-3x - 18=(x - 6)(x + 3)\).

Putting it all together, the denominator factors as:

\((x + 3)(x + 3)(x - 6)(x^2-3x - 18)=(x + 3)^2(x - 6)(x - 6)(x + 3)\)? Wait, no, wait:

Wait, first, we had the original denominator:

\(x^5-3x^4 - 45x^3+27x^2+648x + 972=(x + 3)(x + 3)(x - 6)(x - 6)(x + 3)\)? Wait, no, let's retrace:

First, we found \(x=-3\) is a root, divided to get \(x^4-6x^3-27x^2+108x + 324\). Then \(x=-3\) is a root again, divided to get \(x^3-9x^2+108\). Then \(x = 6\) is a root, divided to get \(x^2-3x - 18\), which factors to \((x - 6)(x + 3)\). So:

Denominator: \((x + 3)(x + 3)(x - 6)(x - 6)(x + 3)\)? Wait, no:

Wait, first factor: \(x + 3\) (from first division)

Second factor: \(x + 3\) (from second division)

Third factor: \(x - 6\) (from third division)

Fourth factor: \(x - 6\) (from factoring \(x^2-3x - 18\))

Fifth factor: \(x + 3\) (from factoring \(x^2-3x - 18\))

Wait, no, let's do it step by step:

Original denominator: \(x^5-3x^4 - 45x^3+27x^2+648x + 972\)

First, factor out \((x + 3)\) twice? Wait, no, first division: divided by \(x + 3\) to get \(x^4-6x^3-27x^2+108x + 324\). Then divided that by \(x + 3\) to get \(x^3-9x^2+108\). Then divided that by \(x - 6\) to get \(x^2-3x - 18\), which factors to \((x - 6)(x + 3)\). So:

Denominator = \((x + 3)(x + 3)(x - 6)(x - 6)(x + 3)\)? Wait, no:

Wait, \((x + 3)\) (first factor) * \((x + 3)\) (second factor) * \((x - 6)\) (third factor) * \((x - 6)\) (fourth factor) * \((x + 3)\) (fifth factor)? No, that would be \((x + 3)^3(x - 6)^2\). Let's check by expanding \((x + 3)^3(x - 6)^2\):

\((x^3 + 9x^2 + 27x + 27)(x^2 - 12x + 36)\)

Multiply term by term:

\(x^3(x^2 - 12x + 36)=x^5 - 12x^4 + 36x^3\)

\(9x^2(x^2 - 12x + 36)=9x^4 - 108x^3 + 324x^2\)

\(27x(x^2 - 12x + 36)=27x^3 - 324x^2 + 972x\)

\(27(x^2 - 12x + 36)=27x^2 - 324x + 972\)

Now, add them all together:

\(x^5 - 12x^4 + 36x^3 + 9x^4 - 108x^3 + 324x^2 + 27x^3 - 324x^2 + 972x + 27x^2 - 324x + 972\)

Combine like terms:

\(x^5 + (-12x^4 + 9x^4) + (36x^3 - 108x^3 + 27x^3) + (324x^2 - 324x^2 + 27x^2) + (972x - 324x) + 972\)

\(x^5 - 3x^4 - 45x^3 + 27x^2 + 648x + 972\). Yes! So the denominator factors to \((x + 3)^3(x - 6)^2\).

Step3: Simplify the Function

Now, the function \(f(x)=\frac{(x + 7)(x - 6)}{(x + 3)^3(x - 6)^2}\). We can cancel out the common factor \((x - 6)\) (assuming \(x eq6\)):

\(f(x)=\frac{x + 7}{(x + 3)^3(x - 6)}\) (for \(x eq6\) and \(x eq - 3\))

Answer:

\(f(x)=\frac{x + 7}{(x + 3)^3(x - 6)}\) (with \(x
eq - 3,6\))