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Algebraic Identities and Polynomials questions for IPMAT, with solutions

10 free Algebraic Identities and Polynomials practice questions for IPMAT Indore, Rohtak and JIPMAT, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Algebraic Identities and Polynomials on Learn.

1. Question 1Multiple choiceWritten by IPMATwayAlgebraic Identities

   If $x-y=4$ and $xy=5$, then $x^{2}+y^{2}$ equals

   1. A.$26$
   2. B.$21$
   3. C.$6$
   4. D.$16$
   Show the answer and solution

   Answer: A. $26$

   Use $x^2+y^2=(x-y)^2+2xy$. Substitute: $4^2+2\cdot5=16+10=26$, option A.

2. Question 2Type the answerWritten by IPMATwayMaxima and Minima in Algebra

   The minimum value of $x^{2}+8x+20$, where $x$ is real, is

   Show the answer and solution

   Answer: 4

   Complete the square: $x^2+8x+20=(x+4)^2+4$. The square is at least $0$, so the minimum is $4$ at $x=-4$.

3. Question 3Type the answerWritten by IPMATwayMaxima and Minima in Algebra

   The minimum value of $p+\frac{1}{p}$ for $p>0$ is

   Show the answer and solution

   Answer: 2

   By AM-GM, $p+\frac1p\ge2\sqrt{p\cdot\frac1p}=2$, with equality at $p=1$. Minimum is $2$.

4. Question 4Type the answerWritten by IPMATwayAlgebraic Identities

   If $x=101$, then the value of $x(x^{2}-3x+3)$ is

   Show the answer and solution

   Answer: 1000001

   Note $x(x^2-3x+3)=x^3-3x^2+3x=(x-1)^3+1$. With $x=101$: $100^3+1=1000001$.

5. Question 5Type the answerWritten by IPMATwayMaxima and Minima in Algebra

   If $a>0$, $b>0$ and $ab=1$, then the least value of $(1+a)(1+b)$ is

   Show the answer and solution

   Answer: 4

   Expanding, $(1+a)(1+b)=1+a+b+ab=2+a+b$. By AM-GM $a+b\ge2\sqrt{ab}=2$, so the least value is $4$ (at $a=b=1$).

6. Question 6Multiple choiceWritten by IPMATwayPolynomials and Remainder Theorem

   The polynomial $x^{29}-x^{25}+x^{13}-1$ is divisible by

   1. A.both $x-1$ and $x+1$
   2. B.$x-1$ but not $x+1$
   3. C.$x+1$ but not $x-1$
   4. D.neither $x-1$ nor $x+1$
   Show the answer and solution

   Answer: B. $x-1$ but not $x+1$

   By the factor theorem test $x=1$ and $x=-1$: $f(1)=0$ but $f(-1)=-2\neq0$. So $x-1$ is a factor and $x+1$ is not.

7. Question 7Multiple choiceWritten by IPMATwayPolynomials and Remainder Theorem

   The GCD of $x^{4}-4x^{2}+3$ and $x^{4}-x^{2}-6$ is

   1. A.$x^{2}+2$
   2. B.$x-1$
   3. C.$x^{2}-1$
   4. D.$x^{2}-3$
   Show the answer and solution

   Answer: D. $x^{2}-3$

   Treat both as quadratics in $x^{2}$: $(x^{2}-1)(x^{2}-3)$ and $(x^{2}-3)(x^{2}+2)$. The common factor is $x^{2}-3$.

8. Question 8Multiple choiceWritten by IPMATwayAlgebraic Identities

   If $a^{1/3}+b^{1/3}+c^{1/3}=0$, then

   1. A.$a+b+c=0$
   2. B.$(a+b+c)^{3}=27abc$
   3. C.$a+b+c=3abc$
   4. D.$(a+b+c)^{3}=3abc$
   Show the answer and solution

   Answer: B. $(a+b+c)^{3}=27abc$

   Put $u=a^{1/3}$, $v=b^{1/3}$, $w=c^{1/3}$. Since $u+v+w=0$, $u^{3}+v^{3}+w^{3}=3uvw$, i.e. $a+b+c=3(abc)^{1/3}$. Cubing both sides, $(a+b+c)^{3}=27abc$.

9. Question 9Multiple choiceWritten by IPMATwayAlgebraic Identities

   The value of $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{4^{2}}\right)\cdots\left(1-\frac{1}{20^{2}}\right)$ is

   1. A.$\frac{19}{40}$
   2. B.$\frac{21}{20}$
   3. C.$\frac{21}{40}$
   4. D.$\frac{1}{20}$
   Show the answer and solution

   Answer: C. $\frac{21}{40}$

   Each factor is $\frac{(k-1)(k+1)}{k^{2}}$. The product of the $\frac{k-1}{k}$ parts for $k=2,\ldots,20$ is $\frac{1}{20}$, and of the $\frac{k+1}{k}$ parts is $\frac{21}{2}$. Hence the value is $\frac{21}{40}$.

10. Question 10Multiple choiceWritten by IPMATwayPolynomials and Remainder Theorem

    If $x^{3}+ax^{2}+bx+6=(x-1)(x-2)(x-c)$ for all $x$, then $a+b$ equals

    1. A.$-13$
    2. B.$7$
    3. C.$-7$
    4. D.$-6$
    Show the answer and solution

    Answer: C. $-7$

    Comparing constant terms, $(-1)(-2)(-c)=-2c=6$, so $c=-3$. Expanding $(x-1)(x-2)(x+3)=x^{3}+0x^{2}-7x+6$, so $a=0$, $b=-7$ and $a+b=-7$.