Class 9 Maths Chapter 2 HOTS Questions - Polynomials

Q.1. If  p(x) = x² – 2√2 x + 1, then find  p (2√2) .

Sol. Since, p(x) = x² – 2 √2 x + 1 , replacing value of x=2√2

= 4 (2) – 4 (2) + 1

= 8 – 8 + 1 

= 1

Q.2. Factorize 3x²-14x - 5$2x^2\; +\; 7x\; +\; 3$using the middle term splitting method.

Sol.$a=3,b=-14,c=-\mathrm{5, }$We need two numbers that multiply to $ac=3\times (-5)=-15ac\; =\; 3\; \backslash times\; (-5)\; =\; -15$ and add up to $b=-14.=\; -14$

The numbers are -15 and 1 because $-15\times 1=-15-15\; \backslash times\; 1\; =\; -15$ and $-15+1=-14.-15\; +\; 1\; =\; -14$

Rewrite the middle term: $3x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}-15x+x-5$

$-\; 15x\; +\; x\; -\; 5$Factor by grouping:$(3x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}-15x)+(x-5)(3x^2\; -\; 15x)\; +\; (x\; -\; 5)$  ⇒ $3x(x-5)+1(x-5)3x(x\; -\; 5)\; +\; 1(x\; -\; 5)$  ⇒ $(3x+1)(x-5)(3x\; +\; 1)(x\; -\; 5)$(3x+1)(x−5)

So, the factorized form of $3x\mathrm{<sup\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sup>}-14x-53x^2\; -\; 14x\; -\; 5$  is $(3x+1)(x-5)(3x\; +\; 1)(x\; -\; 5)$.

Q.3.Evaluate (102) ³  using suitable identity 

Sol. We can write 102 as 100+2
Using identity, (x+y) ³ = x ³ +y ³ +3xy(x+y)
(100+2) ³ =(100) ³ +2 ³ +(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

Q.4.Factorise 4x² – 12x + 9 

Sol:To factorize this expression, we need to find two numbers α and β such that α + β = –12 and αβ = 36

4x² – 6x – 6x + 9

2x(x – 3) – 3(x – 3)

(2x – 3)(x – 3)

Q.5.Using the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case

p(x) = 2x³+x²–2x–1, g(x) = x+1

Sol. Given:p(x) = 2x³+x²–2x–1,  g(x) = x+1

g(x) = 0

⇒ x+1 = 0

⇒ x = −1

∴Zero of g(x) is -1.

Now,

p(−1) = 2(−1)³+(−1)²–2(−1)–1

= −2+1+2−1

= 0

∴By the given factor theorem, g(x) is a factor of p(x).

Q.6.Expanding each of the following, using all the suitable identities:

(i) (x+2y+4z)²

(ii) (2x−y+z)²

Sol:(i) (x+2y+4z)²

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, x = x

y = 2y

z = 4z

(x+2y+4z)² = x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)

= x²+4y²+16z²+4xy+16yz+8xz

(ii) (2x−y+z)²

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, x = 2x

y = −y

z = z

(2x−y+z)² = (2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)

= 4x²+y²+z²–4xy–2yz+4xz

Q.7. Factorize 12x²-29x +15using the middle term splitting method.

Sol.$a=12,b=-29,c=+15,$We need two numbers that multiply to $ac=12\times (15)=\; 180$ and add up to $b=-29.$

The numbers are -20 and -9 because $-20\times -9=180-20\; \backslash times\; -9\; =\; 180$ and $-20+-9=-29-20\; +\; -9\; =\; -29$.

Rewrite the middle term:12x²−20x−9x+15

Factor by grouping:(12x²−20x)+(−9x+15) $\mathrm{\Rightarrow \; 4}x(3x-5)-3(3x-5)4x(3x\; -\; 5)\; -\; 3(3x\; -\; 5)$ ⇒ (4x−3)(3x−5)
So, the factorized form of 12x²−29x+15 is $(4x-3)(3x-5)(4x\; -\; 3)(3x\; -\; 5)$.

Q.8.Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: 25a²–35a+12

(ii) Area: 35y²+13y–12

Sol:(i) Area: 25a²–35a+12

Using the splitting the middle term method,

We have to find a number whose sum = -35 and product =25×12 = 300

We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20 = 300]

25a²–35a+12 = 25a²–15a−20a+12

= 5a(5a–3)–4(5a–3)

= (5a–4)(5a–3)

Possible expression for length  = 5a–4

Possible expression for breadth  = 5a –3

(ii) Area: 35y²+13y–12

Using the splitting the middle term method,

We have to find a number whose sum = 13 and product = 35×-12 = 420

We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420]

35y²+13y–12 = 35y²–15y+28y–12

= 5y(7y–3)+4(7y–3)

= (5y+4)(7y–3)

Possible expression for length  = (5y+4)

Possible expression for breadth  = (7y–3)

Q.10. Factorise 10y² – 28y + 14