Question

according to the rational root theorem, the numbers below are some of the potential roots of \\(f(x) = 10x^3 + 29x^2 - 66x + 27\\). select all that are actual roots.

\\(-\frac{9}{2}\\)

\\(-\frac{9}{10}\\)

\\(\frac{3}{5}\\)

1

3

Explanation:

🆕 New Concept Discovered: Rational Root Theorem and Finding Polynomial Roots
Finding where a polynomial equals zero using factors of the constant and leading coefficient.

Step 1: Test integer candidates

We evaluate the polynomial \( f(x) = 10x^3 + 29x^2 - 66x + 27 \) at the given integer options to see if they yield zero.

For \( x = 1 \):

$$ f(1) = 10(1)^3 + 29(1)^2 - 66(1) + 27 $$

$$ f(1) = 10 + 29 - 66 + 27 = 0 $$

Since \( f(1) = 0 \), \( x = 1 \) is an actual root.

For \( x = 3 \):

$$ f(3) = 10(3)^3 + 29(3)^2 - 66(3) + 27 $$

$$ f(3) = 10(27) + 29(9) - 198 + 27 $$

$$ f(3) = 270 + 261 - 198 + 27 = 360 eq 0 $$

So, \( x = 3 \) is not a root.

Step 2: Factor the polynomial

Since \( x = 1 \) is a root, we can divide \( f(x) \) by \( (x - 1) \) using synthetic division or polynomial long division to find the remaining quadratic factor.

Dividing \( 10x^3 + 29x^2 - 66x + 27 \) by \( x - 1 \):

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$$

The remaining quadratic is \( 10x^2 + 39x - 27 \).

Step 3: Find the remaining roots

Set the quadratic equation to zero to find the other roots:

$$ 10x^2 + 39x - 27 = 0 $$

We can factor this quadratic by grouping. We look for two numbers that multiply to \( 10 \times (-27) = -270 \) and add to \( 39 \). These numbers are \( 45 \) and \( -6 \).

$$ 10x^2 + 45x - 6x - 27 = 0 $$

$$ 5x(2x + 9) - 3(2x + 9) = 0 $$

$$ (5x - 3)(2x + 9) = 0 $$

Solving for \( x \):

$$ 5x - 3 = 0 \implies x = \frac{3}{5} $$

$$ 2x + 9 = 0 \implies x = -\frac{9}{2} $$

Step 4: Compare with the given options

The actual roots of the polynomial are:

$$ x = 1, \quad x = \frac{3}{5}, \quad x = -\frac{9}{2} $$

Comparing this to the given choices:

• \( -\frac{9}{2} \) (Correct)
• \( \frac{9}{10} \) (Incorrect)
• \( \frac{3}{5} \) (Correct)
• \( 1 \) (Correct)
• \( 3 \) (Incorrect)

Answer:

• \( -\frac{9}{2} \)
• \( \frac{3}{5} \)
• \( 1 \)