Question

question 8

given the function (p(n) = -13(n - 5)^8(n + 4)^4(n - 2)),

its roots are

and their multiplicities are .

question 9

find the quotient and remainder using long division for: (\frac{2x^3 - 16x^2 + 7x - 34}{2x^2 + 5})

the quotient is
the remainder is

Explanation:

⚡ Using: Understanding Polynomial Functions · dividing polynomials

Step 1: Solve Question 8 (Roots and Multiplicities)

To find the roots of the polynomial function \( P(n) = -13(n - 5)^8(n + 4)^4(n - 2) \), set each factor equal to zero:

• \( n - 5 = 0 \implies n = 5 \)
• \( n + 4 = 0 \implies n = -4 \)
• \( n - 2 = 0 \implies n = 2 \)

The multiplicity of each root is the exponent of its corresponding factor:

• The root \( n = 5 \) has a multiplicity of \( 8 \).
• The root \( n = -4 \) has a multiplicity of \( 4 \).
• The root \( n = 2 \) has a multiplicity of \( 1 \).

Listing the roots and their corresponding multiplicities in order (e.g., from smallest to largest root):

• Roots: \( -4, 2, 5 \)
• Multiplicities: \( 4, 1, 8 \)

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Step 2: Solve Question 9 (Polynomial Long Division)

Divide \( 2x^3 - 16x^2 + 7x - 34 \) by \( 2x^2 + 5 \):

1. Divide the leading terms:

$$ \frac{2x^3}{2x^2} = x $$

Multiply \( x \) by the divisor:

$$ x(2x^2 + 5) = 2x^3 + 5x $$

Subtract this from the dividend:

$$ (2x^3 - 16x^2 + 7x - 34) - (2x^3 + 5x) = -16x^2 + 2x - 34 $$

1. Divide the new leading term:

$$ \frac{-16x^2}{2x^2} = -8 $$

Multiply \( -8 \) by the divisor:

$$ -8(2x^2 + 5) = -16x^2 - 40 $$

Subtract this from the remaining polynomial:

$$ (-16x^2 + 2x - 34) - (-16x^2 - 40) = 2x + 6 $$

Since the degree of \( 2x + 6 \) is less than the degree of \( 2x^2 + 5 \), the division is complete.

• Quotient: \( x - 8 \)
• Remainder: \( 2x + 6 \)

Answer:

Question 8

• its roots are: \( -4, 2, 5 \) (or in any order: \( 5, -4, 2 \))
• and their multiplicities are: \( 4, 1, 8 \) (corresponding to the order of roots: \( 8, 4, 1 \) for \( 5, -4, 2 \))

Question 9

• The quotient is: \( x - 8 \)
• The remainder is: \( 2x + 6 \)