Polynomials · noteswale

1. Ye topic hai kya?

Ek variable x ke powers ka finite jod — p(x). Paper poochhta: ye polynomial hai ya nahi, degree kya, p(a) kitna, zero kahan, (x−a) factor hai ya remainder kitna, zeros se quadratic banao, cubic toot-na.

Yahan lock

Definition, degree, value, remainder theorem, factor theorem, division algorithm, α+β / αβ, splitting se factor. Graph sirf ek polynomial ke zeros (x-axis cut).

Yahan kya nahi

Do equations, do lines — Pair of linear. x = (−b±√D)/2a, discriminant word-problem — Quadratic equations. Integer HCF Euclid — Real numbers.

2. Polynomial kya hota hai

One variable, real coefficients, exponents whole numbers 0, 1, 2, 3… — negative nahi, fraction nahi, variable root ke andar nahi.

Standard form: p(x) = a_nxⁿ + a_n−1xⁿ⁻¹ + … + a₁x + a₀
a_n ≠ 0 = leading coefficient. a₀ = constant term.
Har a_i real ho sakta — √2, π, −7, 1/3 sab chalenge coefficient mein. Problem tab jab x ke upar galat power ho.

Term = ek tukda jaise −3x². Monomial ek term, binomial do, trinomial teen. Ye names extra hain; degree zyada important.

Expression	Polynomial?	Kyun
4x³ − x + 9	haan	powers 3, 1, 0
√3 x² + 1	haan	√3 coefficient hai, x ke upar 2
π	haan	constant = π · x⁰
√x + 2	nahi	x^1/2 — fraction power
1/x + x	nahi	x⁻¹
(x²+1)/x	nahi	simplify: x + 1/x, wahi negative power
2^x	nahi	variable exponent, polynomial nahi
3/(x−1)	nahi	denominator mein x

√x nahi, lekin √2 · x haan. Root kis cheez pe hai — coefficient ya variable.
x² + 1/2 polynomial hai (½ constant). 1/(2x) nahi.
Do variables x² + y is chapter ka one-variable polynomial nahi — Pair of linear / alag.

Likho hamesha badi power pehle: 2 + 5x − x² ko −x² + 5x + 2. Degree galat na padh jaye.

3. Degree — sabse oonchi zinda power

Degree = highest exponent jiska coefficient zero nahi. 0·x⁵ + 3x² + 1 mein x⁵ mari — degree 2, 5 nahi.

Naam	Degree	Example
Zero polynomial	not defined	p(x)=0 (sab coefficients 0)
Constant (non-zero)	0	7, −3, √2
Linear	1	2x − 5
Quadratic	2	x² − 4x + 3
Cubic	3	x³ − 2x + 1
Biquadratic	4	x⁴ − 1

Zero polynomial alag jaanwar: 0+0x+0x²… koi “highest non-zero” nahi → degree define nahi.
Constant 7 = 7x⁰ → degree 0. 0 aur 7 ko ek mat samajhna.

Jod / guna pe degree

Do non-zero polynomials: deg(p·q) = deg p + deg q. (x²)(x³)=x⁵, 2+3=5.

Sum: deg(p+q) ≤ max(deg p, deg q). Barabar tab jab leading cancel na ho. Cancel example: (x²+x) + (−x²+4) = x+4, max 2 tha, result 1.

Degree of 5? 0, “no x isliye no degree” galat — unless wo 0 khud ho.
3x³ + 2x³ = 5x³, degree 3 rehti, 6 nahi.

4. p(a) aur zero — graph pe x-axis

p(a) = x ki jagah a rakh do. Zero (root) wo number α jahan p(α) = 0. Graph pe yeh x-axis ko kaat-ta / chhoo-ta point.

Linear ax+b=0 (a≠0) → exactly ek zero: x = −b/a.
Degree n (n≥1) ka polynomial: at most n real zeros — unless identically zero polynomial.
Zero ho hi nahi sakte (x²+1), ek ho (touch), n ho. “Hamesha n zeros” galat — wo complex count karke baad ki class.

Linear · p(x)=x−2

Seedhi line. Ek cut, x=2. p(2)=0. p(0)=−2 (y-intercept).

Quadratic · (x−1)(x−3)

Parabola. Do cuts: 1 aur 3. Beech x=2 pe neeche (vertex). x²+1 jaisi upar hi rehti, cut zero.

Do alag lines ka milna yahan nahi — wo Pair of linear. Yahan ek graph, zeros = x-axis.

Repeat zero: (x−2)² = x²−4x+4. Zero sirf 2, lekin do baar (multiplicity 2). Graph x=2 pe touch karta, paar nahi. Count “at most 2” mein yeh 2 hi zeros multiplicity se, visually ek point.

p(a) nikaalo: 2x² − 3x + 1, a=2 → 2·4 − 3·2 + 1 = 8−6+1=3. Zero nahi, remainder theorem mein kaam aayega.

5. Remainder theorem aur factor theorem

Poori division kiye baghair remainder. Yahi LDC ka tez hathyar.

p(x) ko (x − a) se divide: remainder constant hota (degree 0, kyunki divisor degree 1 se chhota).
Remainder = p(a).
Divisor (x + a) = (x − (−a)) → remainder p(−a).

Kyun? Division algorithm: p(x) = (x−a)·q(x) + r. x=a rakho: p(a)=0·q(a)+r → r=p(a). r constant isliye kyunki deg r < 1.

Divisor (ax − b), a≠0: remainder p(b/a). Kyunki ax−b=0 ⇒ x=b/a. Example: (2x−1) se divide, remainder p(1/2).

Example · remainder

p(x)=x³ − 4x + 5 ko (x−2) se divide. Remainder?

Solution

Theorem
x−2 ⇒ a=2. Remainder = p(2). Divide mat karo.
Plug
p(2)=8 − 4·2 + 5 = 8−8+5=5.

Answer5

Factor theorem: (x−a) tab factor jab remainder 0, matlab p(a)=0.
Ulta bhi: p(a)=0 ⇒ (x−a) p(x) ko divide karta, koi remainder nahi.

Zero dhoondhna = linear factor dhoondhna. Cubic pe trial: ±1, ±2, constant ke factors. Jo p(a)=0 kare, (x−a) nikaal, baaki quadratic.

Example · factor?

(x−2), p(x)=x³ − 3x² + 4x − 4 ka factor hai?

Solution

p(2)
8 − 3·4 + 4·2 − 4 = 8−12+8−4=0.
Haan
Remainder 0 → factor. Agar 0 na aata, factor nahi — remainder wahi number hota.

AnswerHaan, factor hai

p(2)=0 matlab (x−2) factor. (x+2) tab jab p(−2)=0.
Remainder 4 aaya to factor nahi — 4 ko “almost zero” mat bolo.

6. Division algorithm

Integers jaisa: dividend = divisor × quotient + remainder, remainder divisor se “chhota”.

Polynomials: p(x) = g(x)·q(x) + r(x)
ya to r(x)=0, ya deg r < deg g.
g(x) ≠ 0. Ye Euclid integer remainder ka polynomial version — HCF(405,126) yahan nahi.

g degree 1 ho to r constant = p(a), upar wala theorem. g degree 2 ho to r linear ya constant: rx + s.

Example · long division

x² + x + 1 ko (x−1) se. Remainder theorem se 3 aana chahiye — division se check.

Solution

Divide

 x + 2 ____________ x − 1 ) x² + x + 1 x² − x ------- 2x + 1 2x − 2 ------- 3

Padho
q(x)=x+2, r=3. Check: (x−1)(x+2)+3 = x²+x−2+3 = x²+x+1. p(1)=3, match.

Answerquotient x+2, remainder 3

LDC mein aksar poori division maangte nahi — remainder / factor kaafi. Cubic factor karte waqt ek linear nikaal ke quadratic divide karni padti hai (neeche factorise).

7. Zeros aur coefficients — quadratic lock, cubic extra

Quadratic ax² + bx + c (a≠0), zeros α, β (real maan ke, ya formal). Expand:

a(x−α)(x−β) = a[x² − (α+β)x + αβ] = ax² + bx + c

Compare: −a(α+β)=b, a·αβ=c.

Quadratic ax²+bx+c:
α + β = −b/a
αβ = c/a
Sign yaad: sum mein minus, product seedha c/a.

x² − 5x + 6: a=1, b=−5, c=6. Sum=5, product=6 → 2 aur 3. Check 2+3=5, 2·3=6.

2x² − 5x + 3: sum=5/2, product=3/2. Zeros 1, 3/2. (1+3/2=5/2, 1·3/2=3/2).

Zeros se polynomial banao

Zeros α, β → p(x) = k(x−α)(x−β), k ≠ 0 koi real.
Paper “a=1” / “monic” / “integer coefficients, smallest” bole to k lock.
Integer chahiye aur zeros fraction: k = denominator. Zeros 1/2, −2 → k=2: 2(x−1/2)(x+2)=(2x−1)(x+2)=2x²+3x−2.

Sum–product se seedha: monic quadratic = x² − (sum)x + (product).

x² + 5x + 6 ke zeros −2, −3 — sum −5 = −b/a, b=+5. Plus minus ulat mat karna.
Doosra zero: α pata, sum S → β = S − α. Product se: β = (c/a)/α.

Example · banao

Zeros 4 aur −1. Monic quadratic?

Solution

Sum / product
4+(−1)=3, 4·(−1)=−4.
Formula
x² − (sum)x + product = x² − 3x + (−4) = x² − 3x − 4.
Check
(x−4)(x+1)=x²−3x−4. p(4)=16−12−4=0, p(−1)=1+3−4=0.

Answerx² − 3x − 4

α²+β², 1/α+1/β — identity, formula nahi

Quadratic formula yahan mat kholo. Zeros ke symmetric sums:

α² + β² = (α+β)² − 2αβ

α²β + αβ² = αβ(α+β)

1/α + 1/β = (α+β)/αβ (αβ≠0, zero 0 na ho)

(α−β)² = (α+β)² − 4αβ

2x² + 5x − 3: sum=−5/2, product=−3/2. α²+β² = 25/4 − 2(−3/2)=25/4+3=37/4.

Cubic — teen zeros

ax³ + bx² + cx + d, zeros α, β, γ:

α+β+γ = −b/a
αβ+βγ+γα = c/a
αβγ = −d/a ← product pe minus, d ke saath.

x³ − 6x² + 11x − 6: a=1, b=−6, c=11, d=−6. Sum=6, pair-sum=11, product=6 → 1,2,3. Check 1+2+3=6, 1·2+2·3+3·1=2+6+3=11, 1·2·3=6.

Ek zero pata ho to factor theorem, divide, baaki quadratic ke α+β, αβ.

Equal zeros / “nature of roots” poora discriminant Quadratic equations page. Yahan itna kaafi: do equal zeros ⇒ (x−α)² factor, graph touch. Real zero na ho (x²+1) — parabola axis nahi kaatti.

8. Factorise — quadratic split, cubic trial

Zeros nikalne ka Class-X polynomial rasta: toot, remainder theorem. Formula (−b±√D)/2a agla chapter.

Quadratic — middle split

ax²+bx+c. Do numbers jin-ka product = ac, sum = b. Phir group.

Example · split

6x² − x − 2 ke zeros.

Solution

ac, b
ac=6·(−2)=−12. Sum −1. Numbers 3, −4. (3+(−4)=−1, 3·(−4)=−12).
Split
6x² + 3x − 4x − 2 = 3x(2x+1) − 2(2x+1) = (3x−2)(2x+1).
Zeros
3x−2=0 → 2/3. 2x+1=0 → −1/2.
Check coeff
Sum=2/3−1/2=4/6−3/6=1/6. −b/a=1/6. Product=(2/3)(−1/2)=−1/3=c/a. Theek.

Answer2/3 aur −1/2

Cubic — ek rational zero trial

Possible simple zeros: constant ke ± factors, leading se. p(1), p(−1), p(2)… jo 0 aaye, (x−a) nikaalo, quadratic divide.

Example · cubic

x³ − 2x² − x + 2 ke saare zeros.

Solution

Trial
p(1)=1−2−1+2=0. (x−1) factor.

Divide by x−1

 x² − x − 2 ________________ x − 1 ) x³ − 2x² − x + 2 x³ − x² -------- −x² − x −x² + x -------- −2x + 2 −2x + 2 -------- 0

Quadratic
x² − x − 2 = (x−2)(x+1). Zeros 2, −1.
Teen
1, 2, −1. Sum=2=−b/a (b=−2). Product=1·2·(−1)=−2=−d/a (d=2, −d=−2).

Answer1, 2, −1

p(1)=0 isliye (x+1) factor — galat. (x−1).
Quadratic split bhool ke “cannot factor” mat likhna agar ac-sum mil jaye.
x²+1 real mein nahi toot-ta — zeros real nahi, polynomial ab bhi valid.

9. Sawal — basic se pro

Q1 · Basic

Kaun sa polynomial hai?

√x + 4
x² + √2
1/x + 3
2^x

Solution

Coefficient vs variable
B: √2 constant coefficient, x² allowed. A: x^1/2. C: x⁻¹. D: exponent variable.

AnswerB

Q2 · Basic

5x³ − 2x + 7 ka degree?

Solution

Highest power
x³ zinda, coefficient 5≠0. Degree 3. 7 constant term hai, degree nahi. 5 leading coeff, degree nahi.

AnswerC · 3

Q3 · Basic

2x + 6 ka zero?

3
−3
6
−6

Solution

Linear
2x+6=0 → x=−3. Check p(−3)=−6+6=0. p(3)=12≠0.

AnswerB · −3

Q4 · Basic · remainder

p(x)=x² − 3x + 2, divisor x−1. Remainder?

0
1
2
−1

Solution

p(1)
1−3+2=0. Remainder 0, isliye (x−1) factor bhi. (x−1)(x−2).

AnswerA · 0

Q5 · Medium

x² − 5x + 6 ke zeros?

−2, −3
2, 3
1, 6
5, 6

Solution

Split
x²−2x−3x+6=(x−2)(x−3). Zeros 2, 3.
Coeff check
Sum 5=−b/a, product 6. A: −2+(−3)=−5, wo x²+5x+6 hota.

AnswerB

Q6 · Medium

Zeros 4 aur −1. Monic quadratic?

x² + 3x − 4
x² − 3x − 4
x² − 3x + 4
x² + 5x − 4

Solution

x² − (sum)x + prod
Sum 3, product −4 → x²−3x−4.
A
+3x matlab sum −3. C: product +4, 4·(−1) nahi.

AnswerB

Q7 · Medium · factor

p(x)=x³ − 3x² + x + 2. (x−2) factor hai?

Haan, p(2)=0
Nahi, p(2)=4
Haan kyunki degree 3
Nahi, p(−2)=0

Solution

p(2)
8 − 3·4 + 2 + 2 = 8−12+2+2=0. Factor theorem: haan.
D
p(−2)= −8 −3·4 −2+2= −8−12=−18 ≠ 0. (x+2) factor nahi. C bakwas.

AnswerA

Q8 · Medium

3x² − 5x + 2. Zeros ka sum aur product?

sum −5, product 2
sum 5/3, product 2/3
sum −5/3, product 2/3
sum 5, product 2

Solution

Formula
a=3, b=−5, c=2. Sum=−b/a=5/3. Product=c/a=2/3.
Trap
C: minus bhool ke b/a. A/D: a se divide nahi kiya. Zeros 1, 2/3: 1+2/3=5/3, 2/3 product.

AnswerB

Q9 · Pro · cubic

x³ − 6x² + 11x − 6 ke zeros?

1, 2, 3
−1, −2, −3
1, −2, 3
2, 2, 2

Solution

p(1)
1−6+11−6=0. (x−1) factor.
Divide / guess
Sum should be 6, product 6. 1,2,3 fit: pair 2+6+3=11. B: sum −6, polynomial x³+6x²+… hota.

AnswerA · 1, 2, 3

Q10 · Pro

p(x)=2x³ + 3x² − 2x − 3. (x+1) se remainder?

0
2
−3
4

Solution

x+1 ⇒ a=−1
p(−1)=2(−1)+3(1)−2(−1)−3= −2+3+2−3=0.
Matlab
Remainder 0, (x+1) factor. 2(−1)³=2(−1)=−2, 3(−1)²=3 — signs carefully.

AnswerA · 0

Practice

Pehle khud, phir neeche kholo.

1 2 3 4 5 6 7 8 9 10

Zero polynomial ka degree?

0
1
Not defined
∞

Solution

Definition
Sab coefficients 0, koi highest zinda power nahi. NCERT/RBSE: not defined. Constant 7 ka degree 0 — alag.

AnswerC

Cubic ke real zeros zyada se zyada?

1
2
3
Infinite

Solution

Degree n
At most n, unless zero polynomial. Cubic → at most 3. Infinite sirf identically 0.

AnswerC · 3

x² + 1 ke real zeros?

1, −1
0
Koi nahi
i, −i (yahan count)

Solution

x²=−1
Real nahi. Graph x-axis nahi kaatti. Complex i Class-X polynomial paper mein “real zeros = none”.

AnswerC

x³ + 1 ko (x+1) se remainder?

1
2
0
−1

Solution

p(−1)
(−1)³+1= −1+1=0. x³+1=(x+1)(x²−x+1).

AnswerC · 0

x² − 7x + k ka ek zero 3. k? Doosra zero?

k=12, doosra 4
k=21, doosra 7
k=10, doosra 3
k=12, doosra 3

Solution

p(3)=0
9 − 21 + k=0 → k=12.
Sum
α+β=7, 3+β=7, β=4. Product 12, 3×4=12 check. D: doosra bhi 3 hota to (x−3)²=x²−6x+9, b=−7 nahi.

AnswerA

Zeros 1/2 aur −2. Integer coefficients, smallest positive leading?

x² + (3/2)x − 1
2x² + 3x − 2
2x² − 3x − 2
x² − 2x − 1/2

Solution

k=2
2(x−1/2)(x+2)=(2x−1)(x+2)=2x²+4x−x−2=2x²+3x−2.
A
Monic fraction coeff — integer nahi. C: 3x ka sign ulta.

AnswerB

2x³ − 3x² − 3x + 2. (x−2) factor?

Haan
Nahi, p(2)=4
Haan p(−2)=0
Cannot say

Solution

p(2)
2·8 − 3·4 − 3·2 + 2 = 16−12−6+2=0. Haan.

AnswerA

2x² + 5x − 3 ke zeros α, β. α² + β²?

25/4
37/4
13/4
−1/4

Solution

Sum, product
α+β=−5/2, αβ=−3/2.
Identity
(α+β)²−2αβ=25/4 − 2(−3/2)=25/4+3=25/4+12/4=37/4. Formula (−b±√D) yahan zaroori nahi.

AnswerB · 37/4

p(x)=(x+3)(2x−1)+7. (x+3) se remainder?

0
7
−7
2x−1

Solution

Algorithm dikha
Pehle se p=(x+3)·(2x−1)+7. Remainder 7, deg 0 < 1.
Theorem
p(−3)=0·(…)+7=7. Factor nahi kyunki 7≠0. D: (2x−1) quotient hai, remainder nahi.

AnswerB · 7

P10

(x−1)(x−2)(x−3) expand ke baad b (x² ka coeff) aur product of zeros?

b=−6, product 6
b=6, product −6
b=−6, product −6
b=11, product 6

Solution

Pehle (x−1)(x−2)
x²−3x+2. ×(x−3)= x³−3x² −3x²+9x +2x−6= x³−6x²+11x−6.
Cubic
b=−6 (x² ka). Product αβγ=1·2·3=6=−d/a, d=−6. D: 11 to c hai (x ka coeff), b nahi.

AnswerA