NCERT Exemplar Solutions: Polynomials - 2 | Mathematics for Grade 9 PDF Download

Exercise 2.4

Q1. If the polynomials az³ + 4 z² + 3 z – 4 and z³ – 4 z + a leave the same remainder when divided by z – 3, find the value of a.

Solution: Let p₁(z) = az³ + 4z² + 3z – 4 and p²(z) = z³ – 4z + 0
When we divide p₁(z) by z – 3, then we get the remainder p,(3).
Now, p₁(3) = a(3)³ + 4(3)² + 3(3) – 4
= 27a + 36 + 9 – 4
= 27a + 41
When we divide p₂(z) by z – 3 then
we get the remainder p₂(3).
Now, p2(3) = (3)³ – 4(3) + a
= 27 – 12 + a
= 15 + a
Both the remainders are same.
p₁(3) = p₂(3)
27a + 41 = 15 + a
27a – a = 15 – 41 .
26a = 26a = – 1

Q2. The polynomial p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.

Solution: Given, p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7
When we divide p(x) by x + 1, then we get the remainder p(– 1).
Now, p(– 1) = (– 1)⁴ – 2(– 1)³ + 3(– 1)² – a(– 1) + 3a – 7
= 1 + 2 + 3 + a + 3a – 7
= 4a – 1
p(– 1) = 19
⇒ 4a – 1 = 19
⇒ 4a = 20
∴ a = 5
∴ Required polynomial = x⁴ – 2x³ + 3x² – 5x + 3(5) – 7   .....[Put a = 5 on p(x)]
= x⁴ – 2x³ + 3x² – 5x + 15 – 7
= x⁴ – 2x³ + 3x² – 5x + 8
When we divide p(x) by x + 2, then we get the remainder p(– 2)
Now, p(– 2) = (– 2)⁴ – 2(– 2)³ + 3(– 2)² – 5(– 2) + 8
= 16 + 16 + 12 + 10 + 8
= 62
Hence, the value of a 5 and remainder is 62.

Q3. If both x – 2 and x – 1/2 are factors of px² + 5x + r, show that p = r.

Solution: As (x - 2)and  (x - 1/2)are the factors of the polynomial px² + 5x + r
i.e., f(2) = 0 and f(1/2) = 0
Now,
f(2) = p(2)² + 5(2) + r = 0
4p + r = -10  .....(1)
And

p/4 + 5/2 + r = 0
p + 10 + 4x = 0

p + 4x = -10     ........(2)
From equation (1) and (2), we get
4p + r = p + 4r
3p = 3x
p = r

Q4. Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.

Solution: Let p(x) = 2x⁴ – 5x³ + 2x² – x + 2 firstly, factorise x² - 3x + 2.
Now, x² - 3x + 2 = x² - 2x - x + 2 [by splitting middle term]
= x(x-2)-1 (x-2)
= (x-1)(x-2)
Hence, 0 of x² - 3x + 2 are land 2.
We have to prove that, 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² - 3x + 2 i.e., to prove that p (1) =0 and p(2) =0
Now, p(1) = 2(1)⁴ – 5(1)³ + 2(1)² -1 + 2
= 2 - 5 + 2 - 1 + 2
= 6 - 6
= 0
and p(2) = 2(2)⁴ – 5(2)³ + 2(2)² – 2 + 2
= 2x¹⁶ - 5x⁸ + 2x⁴ + 0
= 32 – 40 + 8
= 40 – 40
= 0
Hence, p(x) is divisible by x² - 3x + 2.

Q5. Simplify (2x – 5y)³ – (2x + 5y)³.

Solution: (2x -5y)³ – (2x + 5y)³ 
= [(2x)³ – (5y)³ – 3(2x)(5y)(2x – 5y)] -[(2x)³ + (5y)³ + 3(2x)(5y)(2x+5y)]
[using identity, (a – b)³ = a³ -b³ – 3ab  and (a + b)³ = a³ +b³ + 3ab]
= (2x)³ – (5y)³ – 30xy(2x – 5y) – (2x)³ – (5y)³ – 30xy (2x + 5y)
= -2 (5y)³ – 30xy(2x – 5y + 2x + 5y)
= -2 x 125y³ – 30xy(4x)
= -250y³ -120x²y

Q6. Multiply x² + 4y² + z² + 2xy + xz – 2yz by (– z + x – 2 y).

Solution: (x–2y–z)(x² + 4y² + z² + 2xy + xz − 2yz)
= (x − 2y − z)[(x)² + (−2y)² + (−z)² − (x)(−2y) − (−2y)(−z) − (x)(−z)]
= (x)³ + (−2y)³  +(−z)³ – 3(x)(−2y)(−z)
[Using the identity, a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)]
= x³ – 8y³ – z³ – 6xyz

Q7. If a, b, c are all non-zero and a + b + c = 0, prove that

Solution: To prove,
We know that, a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)
= 0(a² + b² + c² – ab – bc – ca) [∵ a + b + c = 0 , given]
= 0
→ a³ + b³ + c³ = 3 abc
On dividing both sides by abc; we get,

Q8. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ –3abc = – 25.

Solution: Given: a + b + c = 5 and ab + bc + ca = 10
We know that: a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
= (a + b + c)[a² + b² + c² – (ab + bc + ca)]
= 5{a² + b² + c² – (ab + bc + ca)} = 5(a² + b² + c² – 10)
Given: a + b + c = 5
Now, squaring both sides, get: (a + b + c)² = 5²
a² + b² + c² + 2(ab + bc + ca)
= 25 a² + b² + c² + 2 × 10
= 25 a² + b² + c² = 25 – 20
= 5
Now, a³ + b³ + c³ – 3abc = 5(a² + b² + c² – 10)
= 5 × (5 – 10)
= 5 × (– 5)
= – 25

Hence proved.

Q9. Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b ) ( b + c) (c + a).

Solution: To prove: (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a)
L.H.S = [(a + b + c)³ – a³] – (b³ + c³)
= (a + b + c – a)[(a + b + c)² + a² + a(a + b + c)] – [(b + c)(b² + c² – bc)]  .......[Using identity, a³ + b³ = (a + b)(a² + b² – ab) and a³ – b³ = (a – b)(a² + b² + ab)]
= (b + c)[a² + b² + c² + 2ab + 2bc + 2ca + a² + a² + ab + ac] – (b + c)(b² + c² – bc)
= (b + c)[b² + c² + 3a² + 3ab + 3ac – b² – c² + 3bc]
= (b + c)[3(a² + ab + ac + bc)]
= 3(b + c)[a(a + b) + c(a + b)]
= 3(b + c)[(a + c)(a + b)]
= 3(a + b)(b + c)(c + a) = R.H.S
Hence proved.