Math, asked by aakritisingh280, 3 months ago

factorise the following​

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Answered by SandeepAW
0

Answer:

e. (a+b)⁴-(a-b)⁴.

(a+b)⁴=a⁴+4a³b+6a²b²+4ab³+b⁴.

(a-b)⁴=a⁴-4a³b+6a²b²-4ab³+b⁴.

(a+b)⁴-(a-b)⁴=(a⁴+4a³b+6a²b²+4ab³+b⁴)-(a⁴-4a³b+6a²b²-4ab³+b⁴).

(a+b)⁴-(a-b)⁴=a⁴+4a³b+6a²b²+4ab³+b⁴-a⁴+4a³b-6a²b²+4ab³-b⁴.

(a+b)⁴-(a-b)⁴=4a³b+4ab³+4a³b+4ab³.

(a+b)⁴-(a-b)⁴=8a³b+8ab³.

f. 16x^5-81x.

x(16x⁴-81).

x(((4x)²)²-(9)²).

x((4x²+9)(4x²-9)).

x((2x)²-(3)²).

x((2x+3)(2x-3)).

x(2x+3)(2x-3).

g. x²+11x+24=0.

x²+8x+3x+24=0.

x(x+8)+3(x+8)=0.

(x+3)(x+8)=0.

x+3=0. & x+8=0.

x=-3. & x=-8.

h. x²+15x+54=0.

x²+9x+6x+54=0.

x(x+9)+6(x+9)=0.

(x+6)(x+9)=0.

x+6=0. & x+9=0.

x=-6. & x=-9.

i. x²+20x+84=0.

x²+14x+6x+84=0.

x(x+14)+6(x+14)=0.

(x+6)(x+14)=0.

x+6=0. & x+14=0.

x=-6. & x=-14.

j. x²-17x+72.

x²-9x-8x+72=0.

x(x-9)-8(x-9)=0.

(x-8)(x-9)=0.

x-8=0. & x-9=0.

x=8. & x=9.

I think this is your answer.

Answered by MrImpeccable
4

ANSWER e):

(a+b)^4-(a-b)^4

We can write it as,

:\implies[(a+b)^2]^2-[(a-b)^2]^2

We know that,

⇒ x² - y² = (x - y)(x + y)

So,

:\implies[(a+b)^2-(a-b)^2][(a+b)^2+(a-b)^2]

Similarly,

:\implies[(a+b+a-b)(a+b-a+b)][(a+b)^2+(a-b)^2]

:\implies[(2a)(2b)][(a+b)^2+(a-b)^2]

:\implies4ab[(a+b)^2+(a-b)^2]

We know that,

⇒ (x ± y)² = x² + y² ± 2xy

So,

:\implies4ab[(a^2+b^2+2ab)+(a^2+b^2-2ab)]

Now,

:\implies4ab[a^2+b^2+2ab+a^2+b^2-2ab]

On simplifying,

:\implies4ab[2a^2+2b^2]

Taking 2 common,

:\implies\bf8ab(a^2+b^2)

OR

:\implies\bf8a^3b+8ab^3

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ANSWER f):

16x^5-81x

Taking x common,

:\implies x(16x^4-81)

We can write it as,

:\implies x(4^2x^{2^2}-9^2)

:\implies x((4x^2)^2-9^2)

So,

:\implies x((4x^2+9)(4x^2-9)

Also,

:\implies x((4x^2+9)((2x)^2-3^2)

Hence,

:\implies x((4x^2+9)(2x+3)(2x-3)

Therefore,

:\implies\bf x(2x-3)(2x+3)(4x^2+9)

\\

ANSWER g):

x^2+11x+24

On splitting the middle term,

:\implies x^2+8x+3x+24

Taking common,

:\implies x(x+8)+3(x+8)

Taking (x + 8) common,

:\implies\bf(x+8)(x+3)

\\

ANSWER h):

x^2+15x+54

On splitting the middle term,

:\implies x^2+6x+9x+54

Taking common,

:\implies x(x+6)+9(x+6)

Taking (x + 6) common,

:\implies\bf(x+9)(x+6)

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ANSWER i):

x^2+20x+84

On splitting the middle term,

:\implies x^2+14x+6x+84

Taking common,

:\implies x(x+14)+6(x+14)

Taking (x + 14) common,

:\implies\bf(x+14)(x+6)

\\

ANSWER j):

x^2-17x+72

On splitting the middle term,

:\implies x^2-8x-9x+72

Taking common,

:\implies x(x-8)-9(x-8)

Taking (x - 8) common,

:\implies\bf(x-8)(x-9)

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