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 - 5using the middle term splitting method.
Sol. We need two numbers that multiply to and add up to
The numbers are -15 and 1 because and
Rewrite the middle term:
Factor by grouping: ⇒ ⇒ (3x+1)(x−5)
So, the factorized form of is .
Q.3. If a + b + c = 9, and ab + bc + ca = 26, find a2 + b2 + c2
Sol.nGiven: a + b + c = 9, ab + bc + ca = 26
Squaring, we get
(a+b+c)2=(9)2
a2+b2+c2+2(ab+bc+ca)=81
⇒ a2+b2+c2+2×26=81
⇒ a2+b2+c2=81−52 =29
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. If a, b, c are all non-zero and a + b + c = 0, prove that
Sol. ∵ a + b + c = 0 ⇒ a3 + b3 + c3 – 3abc = 0
or a3 + b3 + c3 = 3abc
dividing the whole equation by abc, we get
⇒
Q.6. If then find the value of
Sol:
Q.7. Factorize 12x2-29x +15using the middle term splitting method.
Sol. We need two numbers that multiply to and add up to
The numbers are -20 and -9 because and .
Rewrite the middle term: 12x2−20x−9x+15
Factor by grouping: (12x2−20x)+(−9x+15) ⇒ (4x−3)(3x−5)
So, the factorized form of 12x2−29x+15 is .
Q.8. Factorise: (a – b)3 + (b – c)3 + (c – a)3
Q.10. Factorise 10y2 – 28y + 14
Sol: To factorize this expression, we need to find two numbers α and β such that α + β = –28 and αβ = 140
10y2 – 14y – 10y + 14
2y(5y – 7) – 2(5y – 7)
(2y – 2)(5y – 7)
2(y – 1)(5y – 7)
Q.11 Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12.
Sol: Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)2 = 122
=> 9x2 + 12xy + 4y2 = 144
=>9x2 + 4y2 = 144 – 12xy
From the questions, xy = 6
So,
9x2 + 4y2 = 144 – 72
Thus, the value of 9x2 + 4y2 = 72
Q.12.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)
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2. How do you add and subtract polynomials? |
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