IOQM Math Algebra DPP - 1 (pending from q3 onwards)

Q1. Given that

$f(x) = x^4 + 3x^3 + 8x^2 - kx + 11$ is divisible by $x + 3$, find the value of $k$.

Solution:
Put f(-3) = 0.
Solving will give you k = -83/3.

Q2. Given that

$f(x) = x^4 - ax^2 - bx + 2$ is divisible by $(x + 1)(x + 2)$, find the values of $a$ and $b$.
Solution:
f(-1) = 0 and f(-2) = 0
Plug those values and solve.
You would get: a = 6, b = 3

Q3. Given that a polynomial $f(x)$ has remainders 1, 2, 3 when divided by $(x - 1), (x - 2), (x - 3)$, respectively. Find the remainder of $f(x)$ when it is divided by $(x - 1)(x - 2)(x - 3)$.
Solution:
By the remainder theorem:
f(1) = 1, f(2) = 2, f(3) = 3

Now, when we divide f(x) by (x-1)(x-2)(x-3) remainder's degree will be 1 less than divisor.
=> Remainder = ax^2 + bx + c
f(x) = Q(x).(x-1).(x-2).(x-3) + ax^2 + bx + c
f(1) = 1 = a + b + c
f(2) = 2 = 4a + 2b + c
f(3) = 3 = 9a + 3b + c
Solving these, we get: a = 0, b = 1, c = 0
=> R(x) = x.

Q4. 

"If $x^5 - 5qx + 4r$ is divisible by $(x - 2)^2$, find the values of $q$ and $r$."
Solution:
First approach:
x^5 - 5qx + 4r = (x-2)^2.(ax^3 + bx^2 + cx + d)
Now expand the RHS.
And equate the coefficients of same degree terms and solve the resulting equations.
You will get q = 16, r = 32.

Second approach:
Do long division of P(x)  = x^5 - 5qx + 4r by (x-2).

So

$P(x) = (x - 2)Q(x) + (4r + 32 - 10q),$

where
$Q(x) = x^4 + 2x^3 + 4x^2 + 8x + (16 - 5q).$

Divisibility by $(x - 2)$ forces the remainder to vanish:
4r + 32 - 10q = 0

And Q(2) = 0
which gives: 80 - 5q = 0 and q = 16.
Then r = 32.


Q5. Given that 

$f(x)$ is a polynomial of degree 3, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find the f(x).

Solution 1:
Let's say the first divisor leaves the quotient P(x) and second one leaves Q(x).
Let f(x) = a.x^3 + b.x^2 + c.x + d
f(x) = P(x).(x^2 - 1) + 2x - 5
f(x) = Q(x).(x^2 - 4) - 3x + 4
=> f(1) = -3, f(-1) = -7, f(2) = -2, f(-2) = 10

Step 3 — Convert those four function values into linear equations

$$\begin{cases} A + B + C + D = -3 & (f(1)) \\ -A + B - C + D = -7 & (f(-1)) \\ 8A + 4B + 2C + D = -2 & (f(2)) \\ -8A + 4B - 2C + D = 10 & (f(-2)) \\ \end{cases}$$

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Solving these you will get A,B,C,D and hence f(x).

Solution 2:

Let's say the first divisor leaves the quotient P(x) and second one leaves Q(x).
f(x) = P(x).(x^2 - 1) + 2x - 5
f(x) = Q(x).(x^2 - 4) - 3x + 4

Since f(x) is of degree 3 and divisors are of degree 2 => quotients are of degree 1.
=> P(x) = ax + b, Q(x) = cx + d
f(x)

 = (ax + b)(x^2 - 1) + 2x - 5 = a.x^3 + b.x^2 + x(2-a) - b - 5
=  (cx + d)(x^2 - 4) -3x + 4 = c.x^3 + d.x^2 - x(4c + 3) - 4d +4

Equating coeffiecients of terms of same degrees:
a = c
b = d
a-2 = 3 + 4c
4d - 4 = b + 5
Solving these, you would get:
a = -5/3, b = 3, c = 11/3, d = -8
So f(x) = (-5/3).x^3 + 3.x^2 + (11/3).x - 8