Math, asked by Anonymous, 9 months ago

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Answered by Anonymous
1

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....

hope it will help u so mark my answer into brainleast and please post questions one by one it's my request and don't forget to say thank .

Answered by hisu9876
2

Answer:

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