Detailed Answer

To solve this question we need to use the factor theorem.

\[f(-5) = 0 \]

\[ (-5)^3 + 3(-5)^2 + a(-5) + b = 0 \]

\[ -125 + 75 - 5a + b = 0 \]

\[ -5a + b = 50 \]

Then applying the remainder theorem,

\[ f(5) = 70 \]

\[ (5)^3 + 3(5)^2 + a(5) + b = 70 \]

\[ 125 + 75 + 5a + b = 70 \]

\[ 5a + b = -130 \]

Solving this system of 2 equations we find \( a = -18 \) and \( b = -40 \)

Detailed Answer

a) \( f(2) = 2 \quad \Rightarrow \quad a = 1 \)

b) \( f(1) = 0 \quad \Rightarrow \quad a = -1\)

c) \( f(1) = 10 \quad \Rightarrow \quad a = 9 \)

Detailed Answer

a) Firstly, we can factor \(x^2 - x - 2\) as \((x-2)(x+1)\). Then we can write up the definition of long division, namely:

\[ f(x) = (x-2)(x+1)q(x) + x + 5 \]

We can plug in "convenient" numbers into this:

\[ f(2) = 0 + 7 = 7 \quad \Rightarrow \quad 8a + 4b + 6 = 7\]

\[ f(-1) = 0 + 4 = 4 \quad \Rightarrow \quad -a + b + 1 = 4 \]

Solving these two simultaneous equations we get:

\[ a = -\frac{11}{12} \quad \text{and} \quad b = \frac{25}{12} \]

Detailed Answer

a)

Sum of roots: \( -\frac{a}{1} = 4 \quad \Rightarrow \quad a = -4 \).

Product of roots: \( \frac{d}{1} = 0 \quad \Rightarrow \quad d = 0 \).

We know \( x^2 - 5x + 6 = (x-2)(x-3) \) is a factor, thus 2 and 3 are roots.

\[ f(2) = 0 = 16 + 8a + 4b + 2c + d = -16 + 4b + 2c = 0 \]

\[ f(3) = 0 = 81 + 27a + 9b + 3c + d = -27 + 9b + 3c = 0 \]

Solving these simultaneous equations we get:

\[ b = 1 \quad \text{and} \quad c = 6 \]

b) We know that 0, 2, and 3 are roots, and the sum of all roots is 4. Thus, the final root must be -1.

\[ f(x) = x(x+1)(x-2)(x-3) \]

Detailed Answer

a) Using polynomial division:

\[ x^3 - 4x^2 + 5x - 2 \div (x-2) \]

We apply polynomial division:

Thus, the quotient is: \( x^2 - 2x + 1 \)

Detailed Answer

By the Remainder Theorem:

\[ P(1) = a(1)^3 + b(1)^2 + c(1) + d = 4 \Rightarrow a + b + c + d = 4 \]

\[ P(-2) = a(-2)^3 + b(-2)^2 + c(-2) + d = -3 \Rightarrow -8a + 4b - 2c + d = -3 \]

These give two equations:

\[ a + b + c + d = 4 \]

\[ -8a + 4b - 2c + d = -3 \]

These can be solved along with additional given conditions to determine \(a\), \(b\), \(c\), and \(d\).

Detailed Answer

a) Since \((x-2)\) is a factor of \(f(x)\), we substitute \(x=2\) into \(f(x)\):

\[ 2^4 - 3(2)^3 + a(2)^2 + b(2) + 5 = 0 \]

\[ 16 - 24 + 4a + 2b + 5 = 0 \]

\[ 4a + 2b - 3 = 0 \]

By Vieta’s formulas, the sum of the roots of \(f(x)\) is:

\[ \frac{-(-3)}{1} = 3. \]

And the product of the roots (for a quartic polynomial with leading coefficient 1) is:

\[ (-1)^4 \cdot \frac{5}{1} = 5. \]

b) The degree of \(q(x)\) is \(4 - 1 = 3\).

c) Using the formula from part a), we get \(4a + 2b = 3\).

d) We already know one of the roots is 2. Since the sum of the roots of \(f(x)\) is 3, the sum of the roots of \(q(x)\) must be \(3-2=1\). Similarly, the product of the roots of \(f(x)\) is 5, and we already know one of the roots is 2. Thus, the product of the roots of \(q(x)\) must be \(\frac{5}{2}\).

Detailed Answer

a) \((x^2-2x-3) = (x-3)(x+1)\). Thus, plugging 3 into the polynomial should give 0.

\[ 27 + 9(a+1) + 3(2 - 5a) + 1 = 0 \Rightarrow a = \frac{43}{6} \]

b) Sum of the roots is \(-\frac{49}{6}\), so the third root must be \(\frac{37}{6}\).

So, factoring we get \((x-3)(x+1)(x-\frac{37}{6})\).

c) According to the remainder theorem:

\[ f(-5) = (-5 - 3)(-5 + 1)(-5 - \frac{37}{6}) = -\frac{1072}{3} \]