Q.If the sides of a cubic box are increased by 1,2,3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.
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Concept understanding — Polynomial Equations — Basic Definitions and the Quadratic Recap
A polynomial of degree n in x is P(x)=anxn+an−1xn−1+⋯+a1x+a0 with an=0; the corresponding polynomial equation is P(x)=0. A number c with P(c)=0 is called a root (or zero) — the two words describe exactly the same thing. The leading coefficient is an, the leading term is anxn, and a polynomial with an=1 is monic. A polynomial's exponents must be non-negative integers, though its coefficients may be any real or complex number — this is exactly why 3x−1+2, 5x1/2+1, and trigonometric expressions like cosx−sinx are not polynomials, however polynomial-looking they seem.
For the familiar quadratic ax2+bx+c=0 (a=0), the discriminantΔ=b2−4ac governs the roots via x=2a−b±Δ: with real a,b,c, Δ>0 gives real distinct roots, Δ=0 gives equal real roots, and Δ<0 gives no real roots (a non-real conjugate pair — see Complex Conjugate Root Theorem).
Translating a word problem into a polynomial equation. Many real-world conditions — a box's dimensions and volume, an age or rate relationship — translate directly into a polynomial equation once the unknown is named. E.g. a box with breadth x, length x+6, height x+3 has volume x(x+6)(x+3); requiring this volume to equal a fixed number gives a cubic equation in x, and solving it (checking which root is a physically valid, positive length) answers the real question. The same principle handles factoring shortcuts too — e.g. x3+64=x3+43=(x+4)(x2−4x+16) shows directly that x=−4 is a zero of x3+64, without any trial-and-error. It also underlies a basic but easily-confused fact: if f,g are polynomials of degree m,n respectively, the compositionh(x)=(f∘g)(x)=f(g(x)) has degree mn (multiplied, not added) — substituting a degree-n expression into every power up to xm of f produces a top term of degree m×n.
Let the cube's side be x; the cuboid is x(x+1)(x+2)(x+3)... wait — re-read: sides increase by 1,2,3, so the cuboid is (x+1)(x+2)(x+3). Its excess over x3 is 6x2+11x+6=52, giving x=2.
✓Final answer
The volume of the cuboid is 60 cubic units.
Step 1. Set up the equation. Let the cube's side be x. The cuboid's dimensions are x+1,x+2,x+3, so its volume is (x+1)(x+2)(x+3)=x3+6x2+11x+6.
Step 2. Use the given volume increase. The increase over the cube's volume x3 is 6x2+11x+6, and this equals 52:
6x2+11x+6=52⟹6x2+11x−46=0.
Step 3. Solve the quadratic.Δ=112−4(6)(−46)=121+1104=1225=352. So x=12−11±35, giving x=2 or x=−1246 (rejected, since a side length must be positive).
Step 4. Compute the cuboid's volume. With x=2: dimensions 3,4,5, so volume =3×4×5=60. Check: cube volume =8; increase =60−8=52✓.
✓Final answer
The volume of the cuboid is 60 cubic units.
Translate the word problem into a cubic (here reducible to a quadratic) equation, solve, and reject the non-physical root.
Forgetting to reject the negative root for a physical side length
Expanding (x+1)(x+2)(x+3) incorrectly (the correct expansion is x3+6x2+11x+6)
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2026Set ANNUAL1 markMCQ
Q.If f and g are polynomials of degrees m and n respectively and if h(x)=(f∘g)(x), then the degree of h is :
(a) mn
(b) mn
(c) nm
(d) m+n
›Reveal solutionSolution
Composing f of degree m with g of degree n produces a leading term xmn, so the degree multiplies rather than adds.
Let f(x)=amxm+⋯ (leading term amxm, am=0) and g(x)=bnxn+⋯ (leading term bnxn, bn=0).
h(x)=(f∘g)(x)=f(g(x))=am(g(x))m+⋯.
The leading term of (g(x))m is (bnxn)m=bnmxnm, which is non-zero, so this is genuinely the leading term of h.
Hence degh=nm=mn.
✓Final answer
(b) mn
CBSE 2024Set ANNUAL1 markMCQ
Q.A zero of x3+64 is :
(a) 4i
(b) 0
(c) −4
(d) 4
›Reveal solutionSolution
The equation has one real root (the real cube root of −64) and two complex roots; checking each option against x3=−64.
x3+64=0⇒x3=−64.
Testing x=−4: (−4)3=−64. This satisfies the equation.
Testing x=4i: (4i)3=64i3=64(−i)=−64i=−64. Not a zero.
Testing x=0: 03=0=−64. Not a zero. Testing x=4: 43=64=−64. Not a zero.
So the real zero among the options is x=−4.
✓Final answer
(c) −4
CBSE 2020Set ANNUAL1 markMCQ
Q.A polynomial equation of degree n always has :
(a) exactly n roots
(b) n distinct roots
(c) n real roots
(d) n imaginary roots
›Reveal solutionSolution
The Fundamental Theorem of Algebra guarantees exactly n roots (counted with multiplicity, in C) for a degree-n polynomial equation — not necessarily real or distinct.
A polynomial equation of degree n has the form a0xn+a1xn−1+⋯+an=0, with a0=0.
The Fundamental Theorem of Algebra states that such an equation has at least one root in C.
Applying this repeatedly (factoring out each root) shows the polynomial factors completely into n linear factors over C, so it has exactly n roots when multiplicities are counted.
These n roots need NOT all be distinct — a root can repeat (multiplicity >1), so "n distinct roots" is not guaranteed.
These n roots need NOT all be real — some or all can be non-real complex numbers (e.g. x2+1=0 has 2 roots, both imaginary), so "n real roots" is not guaranteed.
They also need NOT all be imaginary — e.g. x2−1=0 has 2 real roots — so "n imaginary roots" is not guaranteed either.
The only statement that is always true is that a degree-n polynomial equation has exactly n roots (with multiplicity, over C).
✓Final answer
A polynomial equation of degree n always has exactly n roots — option (a).