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source: ncert_exercises
topic: algebra
class_level: class_9
chapter: ch2
exercise: ex2.4
difficulty: intermediate
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Class 9 | Ch2: Polynomials | Exercise 2.4
Factor Theorem: (x-a) is factor of p(x) if and only if p(a)=0.
Key Identities:
(x+y+z)² = x²+y²+z²+2xy+2yz+2zx
(x+y)³ = x³+y³+3xy(x+y)
(x-y)³ = x³-y³-3xy(x-y)
x³+y³+z³-3xyz = (x+y+z)(x²+y²+z²-xy-yz-zx)
Q1. Determine which has (x+1) as factor:
(i) x³+x²+x+1 (ii) x⁴+x³+x²+x+1 (iii) x⁴+3x³+3x²+x+1 (iv) x³-x²-(2+√2)x+√2
Answer: Zero of x+1 is x=-1.
(i) p(-1)=-1+1-1+1=0. YES factor. ✅
(ii) p(-1)=1-1+1-1+1=1≠0. NOT factor. ✅
(iii) p(-1)=1-3+3-1+1=1≠0. NOT factor. ✅
(iv) p(-1)=-1-1+(2+√2)+√2=2√2≠0. NOT factor. ✅
Q2. Use factor theorem to determine if g(x) is factor of p(x):
(i) p(x)=2x³+x²-2x-1, g(x)=x+1
(ii) p(x)=x³+3x²+3x+1, g(x)=x+2
(iii) p(x)=x³-4x²+x+6, g(x)=x-3
Answer:
(i) p(-1)=-2+1+2-1=0. YES factor. ✅
(ii) p(-2)=-8+12-6+1=-1≠0. NOT factor. ✅
(iii) p(3)=27-36+3+6=0. YES factor. ✅
Q3. Find value of k if x-1 is factor of p(x):
(i) p(x)=x²+x+k (ii) p(x)=2x²+kx+√2 (iii) p(x)=kx²-√2x+1 (iv) p(x)=kx²-3x+k
Answer: p(1)=0 for all.
(i) 1+1+k=0 → k=-2. ✅
(ii) 2+k+√2=0 → k=-(2+√2). ✅
(iii) k-√2+1=0 → k=√2-1. ✅
(iv) k-3+k=0 → 2k=3 → k=3/2. ✅
Q4. Factorise: (i) 12x²-7x+1 (ii) 2x²+7x+3 (iii) 6x²+5x-6 (iv) 3x²-x-4
Answer:
(i) 12x²-7x+1=(4x-1)(3x-1). Split: -7x=-4x-3x. ✅
(ii) 2x²+7x+3=(2x+1)(x+3). Split: 7x=6x+x. ✅
(iii) 6x²+5x-6=(3x-2)(2x+3). Split: 5x=9x-4x. ✅
(iv) 3x²-x-4=(3x-4)(x+1). Split: -x=3x-4x. ✅
Q5. Factorise: (i) x³-2x²-x+2 (ii) x³-3x²-9x-5 (iii) x³+13x²+32x+20 (iv) 2y³+y²-2y-1
Answer:
(i) x³-2x²-x+2=(x-1)(x+1)(x-2). Factor: p(1)=0, p(-1)=0, p(2)=0. ✅
(ii) x³-3x²-9x-5=(x+1)²(x-5). p(-1)=0 twice. ✅
(iii) x³+13x²+32x+20=(x+1)(x+2)(x+10). p(-1)=0, p(-2)=0, p(-10)=0. ✅
(iv) 2y³+y²-2y-1=(y-1)(y+1)(2y+1). p(1)=0, p(-1)=0, p(-1/2)=0. ✅
Q6-Q16: Use identities to expand/factorise (key ones below):
(x+y+z)²=x²+y²+z²+2xy+2yz+2zx.
(x+y)³=x³+3x²y+3xy²+y³. (x-y)³=x³-3x²y+3xy²-y³.
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx).
Special case: if x+y+z=0 then x³+y³+z³=3xyz. ✅ |