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Answers
Solution :
1. Factorise : 125x³ - 64y³
>> (5x)³ - (4y)³
>> ( 5x - 4y)( 25x² + 20xy + 16y²)
2. Find the value of :
(x+y)² + (x-y)²
>> x² + 2xy + y² + x² - 2xy + y²
>> 2(x² + y²)
3. Find the value of m if x+4 is a factor of x² + 3x + m .
If x+4 is a factor then f(-4) = 0
>> (-4)² + 3(-4) + m = 0
>> 16 - 12 + m = 0
>> m = -4
4. Expand (y-√3)²
>> y² - 2√3y + 3
5. If x+1/x = 7 find x³ + 1/x³
(x+1/x)³ = 7³ = 343
>> x³ + 1/x³ + 3(x+1/x) = 343
>> x³ + 1/x³ + 21 = 343
>> x³ + 1/x³ = 322
6. Expand (x-2y-3z)²
>> x² + 4y² + 9z² - 4xy - 12yz - 6xz
7. If 3x+2y = 12 and xy = 6, find the value of 27x³ + 8y³
>> (3x+2y)³ = 12³ = 1728
>> 27x³ + 8y³ + 18xy(3x+2y) = 1728
>> 27x³ + 8y³ + 18×6×12 = 1728
>> 27x³ + 8y³ = 1728 - 1296 = 432
8. Factorise 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
>> (2x + 3y - 4z)²
9. Show that (x-1) is a factor of x¹⁰ - 1 and x¹¹ - 1
>> Show that for f(1), both equal to 0 & hence is a factor
x¹⁰ - 1 = (x-1)(x+1)(x⁴+x³+x²+1)(x⁴-x³+x²-x+1)
x¹¹ - 1 = (x-1)(x¹⁰+x⁹+x⁸+x⁷+x⁶+x⁵+x⁴+x³+x²+1)
10. Find the value of a if (x-a) is a factor of x³ - a²x + x + 2.
>> Let f(x) = x³ - a²x + x + 2
For f(a) = 0 ( as it's a factor)
>> a³ - a³ + a + 2 = 0
>> a = -2
11. For what values of a is 2x³ + ax² + 11x + a + 3 divisible by 2x-1 ?
For this expression to be divisible by 2x-1, f(½) = 0
>> 2(½)³ + a(½)² + 11(½) + a + 3 = 0
>> ¼ + ¼a + 11/2 + a + 3 = 0
>> a(1+¼) + 35/4 = 0
>> (5/4)a = (-35/4)
>> a = -7
12. Factorise 1-2ab-(a²+b²)
>> 1-2ab-a²-b²
>> 1-(a²+2ab+b²)
>> 1-(a+b)²
>> (1+a+b)(1-a-b)
13. If one zero of the polynomial x² - √3x + 40 is 5, find the other zero?
Let us assume that the other zero is k.
( k + 5) =√3
>> k = √3-5
14. Simplify (x³-4-x+4x²)/(x²+3x-4)
>> [ x²(x+4) - 1(x+4) ]/[ x² + 4x - x - 4]
>> [x²-1][x+4]/[x+4][x-1]
>> x+1
15. Factorise 27x³ - 63x² + 49x - 343/27
>> 1/27(729x³ - 1701x² + 1323x - 343)
>> 1/27( 9x-7)³
>> [ ⅓(9x-7)]³
>> [ 3x - 7/3 ]³
16. Using a suitable identity evaluate (98)³
>> (98)³ = (100-2)³ = (100)³ - 3(100)(2)(100-2) - 2³ = 941192
17. Without actual division, prove that (x²-x-2) divides (2x⁴+x³-5x²-8x-4)
Let us factorise x²-x-2 first
x²-x-2 = x²-2x+x-2 = x(x-2) + 1(x-2) = (x+1)(x-2) .
To prove this, we need to show f(-1)=f(2)=0
f(2) = 2(2)⁴+(2)³-5(2)²-8(2)-4 = 0
f(-1)=2(-1)⁴+(-1)³-5(-1)²-8(-1)-4 = 0
18. If (x+2) and (x-1) are the factors of x³ + 10x² + mx + n, find the values of m and n.
f(-2) = f(1) = 0.
>> -8 + 40 -2m + n = 1+10+m+n = 0
>> 2m - 32 = n & n = -m-11
>> 2m - 32 = -m - 11
>> 3m = 21
>> m = 7 and n = -18
19. If both (x-2) and (x-½) are factors of px²+5x+r then ?
(a) p = r
(b) p + r = 0
(c) 2p + r = 0
(d) 2r + p = 0
f(2) = f(½) = 0
>> 4p + 10 + r = ¼p + 5/2 + r = 0
4p + 10 = -r & ¼p + 5/2 = -r
>> p(4-¼) = 5/2-10
>> 15/4p = -15/2
>> ¼p = -½
>> p = -2 and r = -2
p = r(a)
20. x+1 is a factor of xⁿ +1 if
(a) n is an odd integer
(b) n is an even integer
(c) n is a negative integer
(d) n is a positive integer
It is a factor only if n is odd, n € I [ Application of factor theorem ] (a)
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Answers :
1. ( 5x - 4y)( 25x² + 20xy + 16y²)
2. 2(x² + y²)
3. m = -4
4. y² - 2√3y + 3
5. x³ + 1/x³ = 322
6. x² + 4y² + 9z² - 4xy - 12yz - 6xz
7. 432
8. (2x + 3y - 4z)²
10. a = -2
11. a = -7
12. (1+a+b)(1-a-b)
13. √3-5
14. x+1
15. [ 3x - 7/3 ]³
16. 941192
18. m = 7 and n = -18
19. (a)
20. (a)
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Additional Information :
For integers a, b, c € N, the following identities hold true -
(a + b)² = a² + 2ab + b²
(a + b)² = (a - b)² + 4ab
(a - b)² = a² - 2ab + b²
(a - b)² = (a + b)² - 4ab
a² + b² = (a + b)² - 2ab
a² + b² = (a - b)² + 2ab
2 (a² + b²) = (a + b)² + (a - b)²
4ab = (a + b)² - (a - b)²
ab = {(a + b)/2}² - {(a-b)/2}²
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + b)³ = a³ + 3a²b + 3ab² b³
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - 3a²b + 3ab² - b³
a³ + b³ = (a + b)( a² - ab + b² )
a³ + b³ = (a + b)³ - 3ab( a + b)
a³ - b³ = (a - b)( a² + ab + b²)
a³ - b³ = (a - b)³ + 3ab ( a - b )
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