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\\e0 \\ x=2 \\ \Rightarrow x^3-5x^2-2x+24=8-20-4+24=8\\e0 \\ 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\\e0 \\ 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\\e0 \\ \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-(1)/(2))(x-(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}=(12\pm8)/(8) \\ \Rightarrow x=(20)/(8),x=(4)/(8) \\ \Rightarrow x=(5)/(2),x=(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-(4)/(3))^2 \end{gathered}`

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