Question

Determine roots algebraically by factoring:1. x^3 -5x^2 - 2x +24 = 02. 4x^4 - 4x^3 - 51x^2 = 40-106x3. 9x^4 - 42x^3 = 32x-64x^2

Answer 1

1) According to the rational root theorem, the roots of the function are given by p/q, where p is a factor of a_0 and q is a factor of a_n. In our case,

$a_0=24,a_n=a_3=1$

Thus, the possible roots are

$\text{possible roots}=\mleft\lbrace\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12\pm24\mright\rbrace$

We need to test each option,

$\begin{gathered} x=1 \\ \Rightarrow x^3-5x^2-2x+24=1-5-2+24=18\ne0 \\ x=2 \\ \Rightarrow x^3-5x^2-2x+24=8-20-4+24=8\ne0 \\ x=3 \\ \Rightarrow x^3-5x^2-2x+24=27-45-6+24=0 \end{gathered}$

Thus, a root of the equation is x=3; then,

$\Rightarrow x^3-5x^2-2x+24=(x-3)(x^2-2x-8)$

Finally, we can easily factorize the quadratic term,

$\Rightarrow x^3-5x^2-2x+24=(x-3)(x-4)(x+2)$

The answer to question 1) is (x-3)(x-4)(x+2).

2) Given the initial equation,

$\begin{gathered} 4x^4-4x^3-51x^2=40-106x \\ \Rightarrow4x^4-4x^3-51x^2+106x-40=0 \end{gathered}$

Using the same theorem as in part 1)

The possible roots are

$\text{possible roots}=\mleft\lbrace\pm1,\pm2,\pm4\pm5,\pm8,\pm10,\pm20,\pm40,\ldots\mright\rbrace$

Testing the options,

$\begin{gathered} x=1 \\ \Rightarrow4x^4-4x^3-51x^2+106x-40=4-4-51+106-40=15\ne0 \\ x=2 \\ \Rightarrow4x^4-4x^3-51x^2+106x-40=64-32-204+212-40=0 \end{gathered}$

Then, a root of the equation is x=2; thus,

$4x^4-4x^3-51x^2+106x-40=(x-2)(4x^3+4x^2-43x+20)$

Using the Rational Root theorem on the cubic term,

$\text{possible roots=}\mleft\lbrace\pm1,\pm2,\pm4,\pm10,\pm20,\ldots\mright\rbrace$

Testing each option,

$\begin{gathered} x=1 \\ \Rightarrow4x^3+4x^2-43x+20=4+4-43+20\ne0 \\ \ldots \\ x=-4 \\ \Rightarrow4x^3+4x^2-43x+20=0 \end{gathered}$

Thus,

$\begin{gathered} \Rightarrow4x^4-4x^3-51x^2+106x-40=(x-2)(x+4)(4x^2-12x+5) \\ \Rightarrow4x^4-4x^3-51x^2+106x-40=(x-2)(x+4)(x-\frac{1}{2})(x-\frac{5}{2}) \end{gathered}$

Where

$\begin{gathered} 4x^2-12x+5=0 \\ \Rightarrow x=\frac{-(-12)\pm\sqrt[]{(-12)^2-4\cdot4\cdot5}}{2\cdot4}=\frac{12\pm\sqrt[]{64}}{8}=\frac{12\pm8}{8} \\ \Rightarrow x=\frac{20}{8},x=\frac{4}{8} \\ \Rightarrow x=\frac{5}{2},x=\frac{1}{2} \end{gathered}$

Therefore, the answer to part 2) is (x-2)(x+4)(x-1/2)(x-5/2)

c) Given the equation

$9x^4-42x^3=32x-64x^2$

Then,

$\begin{gathered} \Rightarrow9x^4-42x^3+64x^2-32x=0 \\ \Rightarrow x(9x^3-42x^2+64x^{}-32)=0 \end{gathered}$

Using the Rational root theorem on the cubic part,

$\Rightarrow\text{possible roots}=\mleft\lbrace\pm1,\pm2,\pm3,\pm4,\pm8,\pm16,\pm32,\ldots\mright\rbrace$

Testing each possibility until finding a root,

$\begin{gathered} x=1 \\ \Rightarrow9x^3-42x^2+64x^{}-32=9-42+64-32=-1 \\ x=2 \\ \Rightarrow9x^3-42x^2+64x^{}-32=0 \end{gathered}$

Thus,

$\begin{gathered} \Rightarrow9x^4-42x^3+64x^2-32x=x(x-2)(9x^2-24x+16) \\ \Rightarrow9x^4-42x^3+64x^2-32x=x(x-2)(x-\frac{4}{3})^2 \end{gathered}$

The answer to part c) is x(x-2)(x-4/3)^2