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Answers
1).
We know, (a+b)³ = a³+3a²b+3ab²+b³ --------1).
given expression; 8p³-27q³-3p²q+54pq²= (3p)³+(2q)³+3(3p)²(2q)+3(3p)(2q)² _______2).
by comparing 1). and 2). we can write 8p³-27q³-3p²q+54pq²= (3p+2q)³
2.
Given,
f(x) = 2x³ + x²+bx-6
According to question,
if f(x) i.e 2x³ + x²+bx-6 is divided by x-3 then it leaves a reminder of 36.
so ,
2x³ + x² + bx - 6 - 36 will be divisible by x-3.
so that
(x-3) is an factor of 2x³ + x²+bx-42 = 0. ---------------------(i)
∴ x - 3 =0
x = 3
put the value of x in eqn(i).
2(3)³ + 3²+3b-42 = 0
⇒ 2*27 + 9 + 3b -42 =0
⇒ 54 + 9 -42 + 3b = 0
⇒ 21 = -3b
∴ b = - 21/3 = -7.
Now,
f(x) = 2x³ + x² - 7x - 6 -----------------------------(ii)
put x = 2;
f(x) = 2(2)³ + 2² - 7*2 - 6 = 16+4-14-6 = 0
so the one factor of the f(x) is x-2.
Divide the eqn(ii) by (x-2) to find out remaining two factors.
x-2) 2x³ + x² - 7x - 6 (2x² + 5x +3
-(2x³ - 4x²)
------------------------
0 + 5x² - 7x - 6
- (5x² - 10x)
----------------------
0 + 3x - 6
- (3x - 6)
------------------
0 + 0
Hence another factor will be 2x² + 5x +3 but, it's in a quadratic form so solve this quadratic equation to get the factors.
2x² + 5x + 3 = 0
⇒ 2x² + 3x + 2x +3 = 0
⇒ x(2x +3) +1(2x + 3) = 0
⇒ (x + 1)(2x + 3) = 0
|
x + 1 = 0 | 2x + 3 = 0
x = - 1 | 2x = -3
| x = -3/2
Therefore the factors are 2, -1 & -3/2
And, Value of b = -7
3.
4x²+9y²+16z²+12xy -24yz -16xz
= (2x)²+(3y)²+(4z)²+2(2)(3)xy -2(3)(4)yz -2(2x)(4z)
= (2x + 3y - 4z)(2x + 3y - 4z)
4.
A - b - a³ + b³
= a - b - (a³ - b³)
we know,
A³ - B³ = (A - B)(A² + AB + B²) use this ,
= ( a - b) - (a - b)(a² + ab + b²)
= (a - b){ 1 - (a² + ab + b²) }
=(a - b)( 1 - a² - ab - b²)
5.
Solution:-
given equetion:-
p(x)=x^3-px^2-qx-30
zeroes are = (x-2) and (x+3)
by ,x = 2
(2)^3 -p(2)^2-2q-30=0
8 - 4p -2q-30 =0
4p+2q = -22.........(1)
by, x= -3
(-3)^3 - p(-3)^2-q×(-3)-30 = 0
-27 -9p+3q -30
-3p+q = 19........(2)
muliply by 2 in eq(2) -eq(1)
-6p+2q = 38..........(2)
4p+2q = -22.........(1)
+_-____+______________
-2p = 60
p = -30 ans
put p's value in eq(1)
4(-30) +2q = -22
2q = -22 +120
q = 98
q = 49 ans
(p,q) = (-30, 49) ans
6.
vnmurty
P(a) = (a² - 2 a)² - 23 (a² - 2 a) + 120
= x² - 23 x + 120
x = a² - 2 a
120 = 15 * 8 15 + 8 = 23
then x² - 15 x - 8 x + 120
(x - 15) x - 8 (x - 15)
(x - 15) ( x - 8)
so P(a) = ( a² - 2a -15) (a² - 2a - 8)
= (a - 5) (a + 3) (a - 4) (a + 2)
by factorizing the two factors
7.
Here Your Answer
3x⁴ - 243
= 3( x⁴ - 243)
= 3( x⁴ - 81)
= 3 ( x ⁴ - 34)
Using identity a² - b² = ( a - b ) ( a + b )
= 3 [(x⁴)² - (32)²)
= 3 ( x⁴ - 32 ) ( x⁴ + 32)
8.
Let f(x) = x³ - ax² + 6x - a
So,
------
When,
-----------
f(a) = x³ - ax² + 6x - a
=> (a)³ - a×(a)² + 6×a - a
=> a³ - a³ + 6a - a
=> 5a
Therefore,
------------------
Remainder = 5a
8.
→ p( x ) = x³ - 13x - 12
=> p( -1 ) = ( -1 )³ - 13( -1 ) -12
=> p( -1 ) = -1 + 13 - 12 = 0
=> By Remainder / Factor theorem, ( x - [ -1 ] ) is a factor
=> p( x ) = ( x + 1 )( x² - x - 12 )
♦ p( -3 ) = ( -3 + 1 )( 9 - ( -3 ) - 12 ) = ( -2 )( 0 ) = 0
♦ p( 4 ) = ( 4 + 1 )( 16 - 4 - 12 ) = 5 x 0 = 0
=> p( x ) = ( x + 1 )( x + 3 )( x - 4 )
9.
Answer:
(999)^3=997002999
Step-by-step explanation:
Given : Expression (999)^3
To find : Evaluate expression using suitable identities?
Solution :
We can write the expression as
(999)^3=(1000-1)^3
Now, Expand using identity,
(a-b)^3=a^3-b^3-3ab(a-b)
Substitute, a=1000 and b=1
(1000-1)^3=(1000)^3-1^3-3(1)(1000)(1000-1)
(1000-1)^3=1000000000-1-(3000)(999)
(1000-1)^3=1000000000-1-2997000
(1000-1)^3=997002999
Therefore, (999)^3=997002999
11.
64m³ - 343n³
= (4m)³ - (7n)³
Use, formula,
a³ - b³ =( a - b)(a² + ab + b²)
= (4m - 7n) {(4m)² + (4m)(7n) + (7n)² }
= (4m - 7n) {16m² + 28mn + 49n²}
12.
Let f(x) = 2x³ - 5x² + kx - x + 2.
x-1 is a factor of f(x)
<=> f(1) = 0
<=> 2 - 5 + k - 1 + 2 = 0
<=> k - 2 = 0
<=> k = 2
13. factorize following:-)
a. => x² + 9x + 18
=> x² + 6x + 3x +18
=> x (x+6) + 3 (x+6)
=> (x+6) (x+3)
Therefore, (x+6) (x+3) are the factots....
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Answer:
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