Question

NO LINKS!! URGENT HELP PLEASE!!!

3. Find a factorization of 3x^4 - 8x^3 + 29x^2 + 140x - 300, given that -3 and 2 + 4i are roots.

a. (x - 3)(x^2 + 4x + 20)(3x - 5)
b. (x +3)(x^2 - 4x + 20)
c. 3(x + 3)(x^2 - 4x + 20)
d. (x + 3)(x^2 - 4x + 20)(3x - 5)

Answer 1

`\textsf{d.}\;\;(x + 3)(x^2 - 4x + 20)(3x - 5)`

Explanation:

To find a factorization of the polynomial 3x⁴ - 8x³ + 29x² + 140x - 300, we can start by using the given roots to determine some of the factors.

Given roots:

• -3
• (2 + 4i)

Since complex roots occur in conjugate pairs for polynomials with real coefficients, we also have the root (2 - 4i).

To find the factors corresponding to the roots -3, (2 + 4i), and (2 - 4i), we can use the factor theorem.

The factor theorem states that if a polynomial f(x) has a root r, then (x - r) is a factor of f(x).

Therefore, three factors of the given polynomial are:

• 
  `(x + 3)`

• 
  `(x-(2+4i))=(x-2-4i)`

• 
  `(x-(2-4i))=(x-2+4i)`

Multiply the two complex factors:

`\begin{aligned}(x-2-4i)(x-2+4i)&=x^2-2x+4ix-2x+4-8i-4ix+8i-16i^2\\&=x^2-2x-2x+4ix-4ix-8i+8i+4-16(-1)\\&=x^2-4x+4+16\\&=x^2-4x+20\end{aligned}`

Multiply the product of the two complex factors by the linear factor (x + 3):

`\begin{aligned}(x^2-4x+20)(x+3)&=x^3+3x^2-4x^2-12x+20x+60\\&=x^3-x^2+8x+60\end{aligned}`

As the polynomial we wish to factor is quartic, and the multiplied factors return a cubic polynomial, the remaining factor must be linear.

Perform long division to find the remaining linear factor:

`\large \begin{array}{r}3x-5\phantom{)}\\x^3-x^2+8x+60{\overline{\smash{\big)}\,3x^4 - 8x^3 + 29x^2 + 140x - 300\phantom{)}}}\\{-~\phantom{(}\underline{(3x^4-3x^3+24x^2+180x)\phantom{-b....)}}\\-5x^3+5x^2-40x-300\phantom{)}\\-~\phantom{()}\underline{(-5x^3+5x^2-40x-300)\phantom{}}\\0\phantom{)}\\\end{array}`

Therefore, the three factors are:

• (x + 3)
• (x² - 4x + 20)
• (3x - 5)

So, the fully factored polynomial is:

`\large\boxed{\boxed{(x + 3)(x^2 - 4x + 20)(3x - 5)}}`

Answer 2

`\boxed{\tt d. (x +3)(x^2 - 4x + 20)(3x-5)}`

Given:

The polynomial is
`\tt 3x^4 - 8x^3 + 29x^2 + 140x - 300`

The roots are -3 and 2 + 4i.

Solution:

Let's factorize the given polynomial
`\tt 3x^4 - 8x^3 + 29x^2 + 140x - 300`, we can use the roots provided in the question:

The roots are x = -3, x = 2 + 4i, and x = 2 - 4i.

When a polynomial has a root, its corresponding factor is (x - root). So, for the given roots, the corresponding factors are:

1. For x = -3, the corresponding factor is (x + 3).

2. For x = 2 + 4i, the corresponding factor is (x - (2 + 4i)) = (x - 2 - 4i).

3. For x = 2 - 4i, the corresponding factor is (x - (2 - 4i)) = (x - 2 + 4i).

Now, we can find the factorization:

`\tt 3x^4 - 8x^3 + 29x^2 + 140x - 300 = 3(x + 3)(x - 2 - 4i)(x - 2 + 4i)`

As mentioned earlier, the factor (x - 2 - 4i)(x - 2 + 4i) can be simplified using the difference of squares formula:

`\tt (x - 2 - 4i)(x - 2 + 4i)`

`\tt (x - 2)^2 - (4i)^2`

`\tt (x - 2)^2 - 16i^2`

`\tt (x - 2)^2 - 16(-1) [since\:\: i^2 = -1]`

`\tt (x - 2)^2 + 16`

`\tt x^2-2*2x+4+16`

`\tt x^2-4x +20`

So, the final factorization is:

`\tt 3x^4 - 8x^3 + 29x^2 + 140x - 300 =\underline{ (x + 3)(x^2-4x +20)}`

By multiplying, we get

`\tt x^3-x^2+8x+60`

Find a factorization of
`3x^4 - 8x^3 + 29x^2 + 140x - 300`, given that -3 and 2 + 4i are roots.

since it is a cubic equation but the given expression is in a quadratic equation.

We can find the remaining factor doing long division.

`\begin{array}{r}3x - 5 \phantom{)} \\x^3 - x^2 + 8x + 60 \overline{\smash {\big)} 3x^4 - 8x^3 + 29x^2 + 140x - 300 \phantom{)}} \\{-~\underline{(3x^4 - 3x^3 + 24x^2 + 180x)\phantom{-b....)}}} \\-5x^3 + 5x^2 - 40x - 300 \phantom{)} \\-~\underline{(-5x^3 + 5x^2 - 40x - 300)\phantom{}} \\0 \phantom{)} \\\end{array}`

Therefore, the factors are:
`\tt 3x^4 - 8x^3 + 29x^2 + 140x - 300 =\underline{ (x + 3)(x^2-4x +20)(3x - 5)}`