If $f$ and $g$ are polynomials of degrees $m$ and $n$ respectively, and if $h(x)=(f\circ g)(x)$, then the degree of $h$ is (1) $mn$ (2) $m+n$ (3) $m^n$ (4) $n^m$

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 (1)

Exercise 3.7 · Q2

f

(1)

mn

(2)

m+n

(3)

m^n

(4)

n^m

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✓ Free question

Concept understanding — Polynomial Equations — Basic Definitions and the Quadratic Recap

A polynomial of degree $n$ in $x$ is $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ with $a_n\ne0$ ; 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 $a_n$ , the leading term is $a_nx^n$ , and a polynomial with $a_n=1$ 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 $3x^{-1}+2$ , $5x^{1/2}+1$ , and trigonometric expressions like $\cos x-\sin x$ are not polynomials, however polynomial-looking they seem.

For the familiar quadratic $ax^2+bx+c=0$ ( $a\ne0$ ), the discriminant $\Delta=b^2-4ac$ governs the roots via $x=\dfrac{-b\pm\sqrt\Delta}{2a}$ : with real $a,b,c$ , $\Delta>0$ gives real distinct roots, $\Delta=0$ gives equal real roots, and $\Delta<0$ 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. $x^3+64=x^3+4^3=(x+4)(x^2-4x+16)$ shows directly that $x=-4$ is a zero of $x^3+64$ , 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 composition $h(x)=(f\circ g)(x)=f(g(x))$ has degree $mn$ (multiplied, not added) — substituting a degree- $n$ expression into every power up to $x^m$ of $f$ produces a top term of degree $m\times n$ .

Step 1. Consider the top-degree term of $f$ . $f(x)=a_mx^m+\cdots$ for some $a_m\ne0$ .

Step 2. Substitute $g(x)$ (degree $n$ ) for $x$ . $f(g(x))=a_m\big(g(x)\big)^m+\cdots$ ; the leading term of $g(x)$ raised to the $m$ -th power has degree $n\times m=mn$ , and every other term in the expansion has strictly lower degree.

Step 3. Confirm this is indeed the highest-degree term of the composition. No other term in $f(g(x))$ 's expansion reaches degree $mn$ , since even the top term of $f$ contributes at most $mn$ once $g$ (degree $n$ ) is substituted in.

Step 4. Conclude. $\deg(h)=\deg(f\circ g)=mn$ .

✓Final answer

mn

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$f$
(a) $m^n$
(b) $mn$
(c) $n^m$
(d) $m+n$
›Reveal solutionSolution
Composing $f$ of degree $m$ with $g$ of degree $n$ produces a leading term $x^{mn}$ , so the degree multiplies rather than adds.

Let $f(x)=a_mx^m+\cdots$ (leading term $a_mx^m$ , $a_m\ne0$ ) and $g(x)=b_nx^n+\cdots$ (leading term $b_nx^n$ , $b_n\ne0$ ).

$h(x)=(f\circ g)(x)=f(g(x))=a_m\big(g(x)\big)^m+\cdots$ .

The leading term of $\big(g(x)\big)^m$ is $\big(b_nx^n\big)^m=b_n^mx^{nm}$ , which is non-zero, so this is genuinely the leading term of $h$ .

Hence $\deg h=nm=mn$ .

✓Final answer
$mn$
CBSE 2024Set ANNUAL1 markMCQ
$x^3+64$
(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 $x^3=-64$ .

$x^3+64=0\Rightarrow x^3=-64$ .

Testing $x=-4$ : $(-4)^3=-64$ . This satisfies the equation.

Testing $x=4i$ : $(4i)^3=64i^3=64(-i)=-64i\ne-64$ . Not a zero.

Testing $x=0$ : $0^3=0\ne-64$ . Not a zero. Testing $x=4$ : $4^3=64\ne-64$ . Not a zero.

So the real zero among the options is $x=-4$ .

✓Final answer
$-4$
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 $\mathbb{C}$ ) for a degree- $n$ polynomial equation — not necessarily real or distinct.

A polynomial equation of degree $n$ has the form $a_0x^n+a_1x^{n-1}+\cdots+a_n=0$ , with $a_0\ne0$ .

The Fundamental Theorem of Algebra states that such an equation has at least one root in $\mathbb{C}$ .

Applying this repeatedly (factoring out each root) shows the polynomial factors completely into $n$ linear factors over $\mathbb{C}$ , 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. $x^2+1=0$ has 2 roots, both imaginary), so " $n$ real roots" is not guaranteed.

They also need NOT all be imaginary — e.g. $x^2-1=0$ 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 $\mathbb{C}$ ).

✓Final answer
$n$

←Previous question · 3.9.2.2Q1 — A zero of x³+64 is (1) 0 (2) 4 (3) 4i (4) -4 →Next question · 3.9.2.2Q3 — A polynomial equation in x of degree n always has (1) n distinct roots…

Q.If fff and ggg are polynomials of degrees mmm and nnn respectively, and if h(x)=(f∘g)(x)h(x)=(f\circ g)(x)h(x)=(f∘g)(x), then the degree of hhh is

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Q.If $f$ and $g$ are polynomials of degrees $m$ and $n$ respectively, and if $h(x)=(f\circ g)(x)$ , then the degree of $h$ is

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