Question

When the polynomial f(x)=(p-1)x^3+px^2+qx+r,where p,q and r are constants is divided by (x+2) and (x-1), the remainder are -5 and 4 respectively.if(x+1)is a factor of f(x). Find the values of p, q and r. Hence factorise f(x) completely

Answer 1

Given the polynomial \( f(x) = (p-1)x^3 + px^2 + qx + r \), we are told the following conditions:

1. The remainder when \( f(x) \) is divided by \( x + 2 \) is -5, so \( f(-2) = -5 \).
2. The remainder when \( f(x) \) is divided by \( x - 1 \) is 4, so \( f(1) = 4 \).
3. The polynomial \( f(x) \) has \( x + 1 \) as a factor, which implies that \( f(-1) = 0 \).

Using these conditions, we can set up a system of equations to find \( p \), \( q \), and \( r \).

### Step 1: Calculate \( f(-2) = -5 \)

Substituting \( x = -2 \) into \( f(x) \):

\[
f(-2) = (p-1)(-2)^3 + p(-2)^2 + q(-2) + r
\]
\[
= (p-1)(-8) + p(4) - 2q + r
\]
\[
= -8(p-1) + 4p - 2q + r
\]
\[
= -8p + 8 + 4p - 2q + r
\]
\[
= -4p - 2q + r + 8 = -5
\]

This simplifies to:

\[
-4p - 2q + r + 8 = -5 \implies -4p - 2q + r = -13 \tag{1}
\]

### Step 2: Calculate \( f(1) = 4 \)

Substituting \( x = 1 \) into \( f(x) \):

\[
f(1) = (p-1)(1)^3 + p(1)^2 + q(1) + r
\]
\[
= (p-1)(1) + p(1) + q + r
\]
\[
= p - 1 + p + q + r
\]
\[
= 2p + q + r - 1 = 4
\]

This simplifies to:

\[
2p + q + r - 1 = 4 \implies 2p + q + r = 5 \tag{2}
\]

### Step 3: Calculate \( f(-1) = 0 \)

Substituting \( x = -1 \) into \( f(x) \):

\[
f(-1) = (p-1)(-1)^3 + p(-1)^2 + q(-1) + r
\]
\[
= (p-1)(-1) + p(1) - q + r
\]
\[
= -p + 1 + p - q + r
\]
\[
= 1 - q + r = 0 \implies r - q = -1 \implies r = q - 1 \tag{3}
\]

### Step 4: Substitute \( r \) in equations (1) and (2)

Now we substitute \( r = q - 1 \) into equations (1) and (2).

Substituting into equation (1):

\[
-4p - 2q + (q - 1) = -13
\]
\[
-4p - q - 1 = -13 \implies -4p - q = -12 \implies 4p + q = 12 \tag{4}
\]

Substituting into equation (2):

\[
2p + q + (q - 1) = 5
\]
\[
2p + 2q - 1 = 5 \implies 2p + 2q = 6 \implies p + q = 3 \tag{5}
\]

### Step 5: Solve equations (4) and (5)

From equation (5):

\[
q = 3 - p \tag{6}
\]

Now substitute equation (6) into equation (4):

\[
4p + (3 - p) = 12
\]
\[
4p + 3 - p = 12 \implies 3p + 3 = 12
\]
\[
3p = 9 \implies p = 3
\]

Substituting \( p \) back into equation (6):

\[
q = 3 - 3 = 0
\]

Substituting \( p \) and \( q \) back into equation (3) gives:

\[
r = q - 1 = 0 - 1 = -1
\]

### Final Values
So, we have:

\[
p = 3, \quad q = 0, \quad r = -1
\]

### Step 6: Factorize \( f(x) \)

Now substituting \( p \), \( q \), and \( r \) into \( f(x) \):

\[
f(x) = (3-1)x^3 + 3x^2 + 0 \cdot x - 1 = 2x^3 + 3x^2 - 1
\]

To factor \( f(x) \):

Since \( f(x) \) is divisible by \( x + 1 \), perform synthetic division or polynomial long division by \( x + 1 \):

\[
\text{Synthetic division: Setting } x + 1 = 0 \Rightarrow x = -1:
\]
\[
\begin{array}{r|rrrr}
-1 & 2 & 3 & 0 & -1 \\
& & -2 & -1 & 1 \\
\hline
& 2 & 1 & -1 & 0 \\
\end{array}
\]

This division gives us:

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

Now to factor \( 2x^2 + x - 1 \):

To find the roots use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}
\]

This gives us the roots:

\[
x = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1.
\]

### Final Factorization

The quadratic factor can be factored into:

\[
2x^2 + x - 1 = 2(x - \frac{1}{2})(x + 1) = 2(x + 1)(x - \frac{1}{2})
\]

So the completely factored form of the polynomial \( f(x) \) is:

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

Thus, we conclude that the polynomial can be expressed in its factored form as:

\[
f(x) = 2(x + 1)^2 (x - \frac{1}{2}).
\]