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.
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)