IOQM Math Algebra DPP - 1

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

Q6. Factorize:

$f(x)=x^{3}+7x^{2}+14x+8$ 

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a. Systematic search for rational roots (Rational-Root Theorem)

All possible rational roots are

$$\frac{\text{factors of }r}{\text{factors of the leading coefficient}}.$$

Because the polynomial is monic, that list is just the ± divisors of $r$:

$$\pm1,\;\pm2,\;\pm4,\;\pm8.$$

Evaluate $f(x)$ at each of these (synthetic division makes this a 10-second job per value).
Stop when you hit one that gives zero; that value is a root.

Example
$f(-1)=(-1)^{3}+7(-1)^{2}+14(-1)+8=-1+7-14+8=0$.
So $x=-1$ is a root.

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b. Divide out the discovered linear factor

Use synthetic division or long division to peel off $(x-a)$ once a root $a$ is found.
You are left with a quadratic

$$f(x)=(x-a)(x^{2}+bx+c).$$

Continuing the example, dividing by $x+1$ gives
$x^{2}+6x+8$.

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c. Factor the quadratic (always possible in closed form)

• If it has integer roots, you can usually spot them quickly.

• Otherwise, apply the quadratic formula to get exact (possibly irrational) roots.

For $x^{2}+6x+8$:
discriminant $6^{2}-4\cdot1\cdot8=4$, so the roots are
$-3\pm\sqrt{( -3)^{2}-8}= -2,\,-4$.
Hence the quadratic factors as $(x+2)(x+4)$.

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d. Put it all together

$$f(x)=(x-a)(x+u)(x+v).$$

Final answer for the example:

$$\boxed{\,x^{3}+7x^{2}+14x+8=(x+1)(x+2)(x+4)\,}.$$

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Why this method is “foolproof”

1. Finite checklist: The Rational-Root Theorem turns “guessing” into a finite checklist you can’t miss.

2. Guaranteed finish:

   • If you find at least one rational root, you reduce to a quadratic, which you can always solve exactly.

   • If none of the candidate rational numbers gives 0, you know the cubic has no rational factor—it is irreducible over ℚ and any further factorisation requires real/complex (Cardano’s) roots, but you have certainty that you’re done as far as rational factors are concerned.

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Q7. (CHINA/1995) Given that the sum of squares of roots to the equation
$2x^2 + ax - 2a + 1 = 0$
is $7 \dfrac{1}{4}$, find the value of $a$.

Solution:

Let the two roots of

$$2x^{2}+ax-2a+1=0$$

be $r_{1}$ and $r_{2}$.

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1. Express the needed symmetric sums

$$\begin{aligned} r_{1}+r_{2}&=-\frac{B}{A}=-\frac{a}{2}, \\[4pt] r_{1}r_{2}&=\frac{C}{A}=\frac{-2a+1}{2}. \end{aligned}$$

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2. Write the sum of the squares of the roots

$$r_{1}^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}-2r_{1}r_{2} =\left(-\frac{a}{2}\right)^{2}-2\left(\frac{-2a+1}{2}\right) =\frac{a^{2}}{4}+2a-1.$$

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3. Impose the given condition

$$\frac{a^{2}}{4}+2a-1=\frac{29}{4}\qquad\Bigl(7\dfrac14=\frac{29}{4}\Bigr).$$

Multiply by $4$:

$$a^{2}+8a-4=29\;\;\Longrightarrow\;\;a^{2}+8a-33=0.$$

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4. Solve for $a$

$$a=\frac{-8\pm\sqrt{(-8)^{2}-4(1)(-33)}}{2} =\frac{-8\pm\sqrt{64+132}}{2} =\frac{-8\pm14}{2}.$$ $$\boxed{\,a=3\quad\text{or}\quad a=-11\,}$$

(Both values give real roots and satisfy the required sum-of-squares.)

Q8. Given that 

$\alpha$ and $\beta$ are the real roots of $x^2 - 2x - 1 = 0$, find the value of
$5{\alpha }^4+12{\beta }^3\mathrm{.}$

There’s a neat algebraic shortcut, so you never have to write the actual decimal roots.

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1. Use the fact that every root satisfies its own quadratic

For the equation

$$x^{2}-2x-1=0 \quad\Longrightarrow\quad x^{2}=2x+1,$$

any power $x^{n}$ (with $n\ge 2$) can be reduced to a linear expression in $x$.

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2. Reduce the required powers

For $\alpha$

$$\begin{aligned} \alpha^{2} &= 2\alpha+1,\\[2pt] \alpha^{3} &= \alpha(\alpha^{2})=\alpha(2\alpha+1)=2\alpha^{2}+\alpha =2(2\alpha+1)+\alpha=5\alpha+2,\\[2pt] \alpha^{4} &= \alpha(\alpha^{3})=\alpha(5\alpha+2)=5\alpha^{2}+2\alpha =5(2\alpha+1)+2\alpha=12\alpha+5. \end{aligned}$$

For $\beta$

Exactly the same steps give

$$\beta^{3}=5\beta+2.$$

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3. Plug into the expression

$$\begin{aligned} 5\alpha^{4}+12\beta^{3} &=5(12\alpha+5)+12(5\beta+2)\\[2pt] &=60\alpha+25+60\beta+24\\[2pt] &=60(\alpha+\beta)+49. \end{aligned}$$

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4. Use the Vieta relations

For $x^{2}-2x-1=0$,

$$\alpha+\beta = 2, \qquad \alpha\beta = -1.$$

Hence

$$5\alpha^{4}+12\beta^{3}=60(2)+49=120+49=169.$$

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$$\boxed{169}$$

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Why this works (the “trick” in a sentence): any polynomial in $\alpha$ or $\beta$ can be reduced modulo their minimal polynomial $x^{2}-2x-1$, so high powers collapse down to linear forms that depend only on the simple Vieta sums.

Q9.  (CHINA/1997) Given that 

$\alpha$ and $\beta$ are the real roots of $x^2 + 19x - 97 = 0$, and

$$\frac{1 + \alpha}{1 - \alpha} + \frac{1 + \beta}{1 - \beta} = -\frac{m}{n},$$

where $m$ and $n$ are two relatively prime natural numbers. Find the value of $m + n$.

Solution:

Solution Sketch

1. Roots of the quadratic
   For $x^{2}+19x-97=0$ let the real roots be $\alpha,\beta$.

   $$\alpha+\beta = -19,\qquad \alpha\beta = -97 \quad\text{(Vieta).}$$

2. Target expression
   Define $f(t)=\dfrac{1+t}{1-t}$. We need

   $$E=f(\alpha)+f(\beta)=\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}.$$

3. Combine the two fractions

   $$E=\frac{(1+\alpha)(1-\beta)+(1+\beta)(1-\alpha)}{(1-\alpha)(1-\beta)}.$$

4. Evaluate the numerator

   $$(1+\alpha)(1-\beta)=1-\beta+\alpha-\alpha\beta,\\ (1+\beta)(1-\alpha)=1-\alpha+\beta-\alpha\beta,\\ \text{Sum}=2-2\alpha\beta=2(1-\alpha\beta).$$

5. Evaluate the denominator

   $$(1-\alpha)(1-\beta)=1-(\alpha+\beta)+\alpha\beta =1-(-19)+(-97)=-77.$$

6. Substitute Vieta values

   $$E=\frac{2(1-(-97))}{-77}=\frac{2\cdot98}{-77}=-\frac{196}{77}=-\frac{28}{11}.$$

7. Identify $m,n$ and sum
   Coprime positive integers: $m=28,\;n=11$.

   $$m+n = 28+11 = \boxed{39}.$$

Q10. The product of two of the four roots of the quartic equation

$x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$
is $-32$. Determine the value of $k$.

Solution:

A “build-the-factor” route

Instead of working with all four roots at once, you can manufacture the quadratic that must contain the two special roots and then force the quartic to split into two quadratics.

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1. Give those two roots a home

If the two roots have product $-32$, call them $r$ and $s$.
They satisfy

$$x^{2}-(r+s)x+rs \;=\;0\quad\Longrightarrow\quad x^{2}-Px-32,$$

where $P=r+s$.

So the quartic must factor as

$$\boxed{(x^{2}-Px-32)\,(x^{2}+Ax+B)}.$$

(The second quadratic, with the remaining two roots $t,u$, has unknown coefficients $A=t+u$ and $B=tu$.)

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2. Expand and match coefficients

$$\begin{aligned} (x^{2}-Px-32)(x^{2}+Ax+B) &= x^{4} + (A-P)x^{3} + (B-AP-32)x^{2} \\ &\quad + (-32A-BP)x + 32B. \end{aligned}$$

Compare with

$$x^{4}-18x^{3}+kx^{2}+200x-1984.$$

That gives the system

$$\begin{cases} A-P = -18,\\[4pt] B-AP-32 = k,\\[4pt] -32A-BP = 200,\\[4pt] 32B = -1984. \end{cases}$$

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3. Solve the little system

1. From the constant term
   $32B=-1984 \Rightarrow B= -62.$

2. From the linear term

   $$-32A -(-62)P =200 \;\Longrightarrow\; -32A+62P=200. \tag{★}$$

3. From the cubic term

   $$A-P=-18 \;\Longrightarrow\; A = P-18. \tag{✦}$$

   Insert (✦) into (★):

   $$-32(P-18)+62P = 200 \;\;\Rightarrow\;\; 94P = -376 \;\;\Rightarrow\;\; P = -\frac{188}{47}.$$

4. Then $A = P-18 = -\dfrac{658}{47}.$

5. Finally, from the quadratic-term equation

   $$k = B - AP - 32 = (-62) - \Bigl(-\tfrac{658}{47}\Bigr)\Bigl(-\tfrac{188}{47}\Bigr) - 32.$$

   The product $AP$ works out to $56$, so

   $$k = -62 - 56 - 32 = 86.$$

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Result

A completely different (but equally legitimate) path arrives at the same answer:

$$\boxed{k = 86}.$$

The key idea was to introduce an explicit quadratic factor with the known product $-32$, expand the factorisation, and match coefficients—no symmetric-sum algebra required.

Q11. 

Find all $a, b$, such that the roots of $x^3 + ax^2 + bx - 8 = 0$ are real and in G.P.

Solution:
Let the 3 roots be p,pr,pr^2.
Multiplication of roots  = 8 = p^3.r^3 => pr = 2
=> Second root is 2.
We can rewrite the roots as 2/r,2,2r
Sum of roots = -a = 2(1 + 1/r + r)
Pairwise sum = b = 4(1/r + 1 + r)
=> b = -2a
This gives us the basic condition for generating (a,b) pairs.

Now we can limit the interval of a and hence b.
2 methods to do that:
Method 1:
Divide the original equation by (x-2).
Quotient will be a quadratic.
For the roots of quotient to be real(as stated in the question), its discriminant >= 0.
If you do that you will get (a+6).(a-2) >= 0
Now 2 possibilities:
a+6 >= 0 AND a-2 >= 0 => a is [2, Inf)
a+6 <= 0 AND a-2 <= 0 => a is (-Inf,6]
So answer: a belongs to [2, Inf) U  (-Inf,6]

Method 2:
a = -2(1 + r + 1/r)
Let's focus on 1 + r + 1/r.
By AM GM inequality:
For r > 0
1 + r + 1/r >= 3 
=> 2.(1 + r + 1/r) >= 6
=> -2.(1 + r + 1/r) <= -6
=> a <= -6

For r < 0:
1 + r + 1/r <= -1
=> 2.(1 + r + 1/r) <=  -2
=> -2.(1 + r + 1/r) >= 2
=> a >= 2

So interval for a is (-Inf,-6] U [2, Inf)
And b = -2a.

Q12. Let $p(x)$ be a polynomial with degree 2008 and leading coefficient 1 such that
$p(0) = 2007, \, p(1) = 2006, \, p(2) = 2005, \, \ldots, \, p(2007) = 0$; determine $p(2008)$.
Solution:

Answer:

$$p(2008)=2008!\;-\;1.$$

Why this works (quick outline)

1. You’re told $p(x)$ is degree 2008 with leading coefficient 1 and

   $$p(k)=2007-k \qquad\text{for }k=0,1,\dots,2007.$$

2. Form a new polynomial

   $$q(x)=p(x)+x-2007.$$

   For every $k=0,\dots,2007$:

   $$q(k)=p(k)+k-2007 =(2007-k)+k-2007=0,$$

   so each of those integers is a root of $q$.

3. A monic polynomial whose roots are exactly $0,1,\dots,2007$ is

   $$R(x)=x(x-1)(x-2)\dotsm(x-2007).$$

   Both $q$ and $R$ are monic of degree 2008 and share all 2008 roots, so they must be identical:

   $$q(x)=R(x).$$

4. Hence

   $$p(x)=R(x)-x+2007.$$

5. Evaluate at $x=2008$:

   $$p(2008)=R(2008)-2008+2007=\bigl[(2008)(2007)\dotsm1\bigr]-1 =2008!-1.$$

That’s the whole story—no massive expansion required!

Q13.

If $P(x)$ denotes a polynomial of degree $n$, such that
$P(k) = \frac{1}{k} \text{ for } k = 1, 2, 3, \ldots, n+1,$
determine $P(n+2)$.

Solution:
Now the first instinct is to create a new function:
Q(x) = P(x) - 1/x
But now Q(x) is no longer a polynomial.
Because in a polynomial, all powers of x are positive integers.
So you can modify it a bit like this:
Q(x) = x.P(x) - 1.
Now Q(x) is a polynomial.

Q(1) = 0 = Q(2).... = Q(n+1)
=> Q(x) is a polynomial with n+1 roots.
Q(x) = K.(x-1)(x-2)....(x - (n+1))

To find K, Q(0) = K.(-1)^(n+1).(n+1)! = -1
=> K = (-1)^n.1/(n+1)!
P(n+2) = (Q(n+2) + 1)/(n+2)
={1/(n+2)}.{1+K.(n+1)!}
= (1+(-1)^n)/(n+2)
= 2/(n+2) when n is even 0 otherwise.

Q14. If one root of the equation 

$2x^2 - 6x + k = 0$ is $\frac{1}{2}(a + 5i)$ where $i^2 = -1$; $k, a \in \mathbb{R}$,
find the values of ‘$a$’ and ‘$k$’.

Solution:
Complex roots occur in conjugate pairs.
So the other root is 1/2(a - 5i).
Sum of roots = 3 = a.
k/2 = 1/4(a^2 + 25) = 17/2 => k= 17.
Answer: a = 3, k = 17.

Q15. Show that, if 

$a, b, c$ are real numbers and $ac = 2(b + d)$, then, at least one of the equations
$x^2 + ax + b = 0$ and $x^2 + cx + d = 0$ has real roots.

Solution:
We will prove this by contradiction.
Let's say both the given equations don't have real roots.
So D < 0 for both.
a^2 - 4b < 0 => a^2 < 4b
c^2 - 4d < 0 => c^2 < 4d
Add both:
a^2 + c^2 < 4(b+d)
Use ac = 2(b+d) as given.
a^2 + c^2 < 2ac
(a-c)^2 < 0
Which is not possible.
Hence proved by contradiction.