Question

Factorise:
30-17(x-y)-21(x-y)^2

Do it with full with step by step explanation

Highly qualified answer needed​

Answer 1

Answer:

30−17(x−y)−21(x−y)2

Distribute:

=30+(−17)(x)+(−17)(−y)+−21x2+42xy+−21y2

=30+−17x+17y+−21x2+42xy+−21y

Answer:

=−21x2+42xy−21y2−17x+17y+30

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:30 - 17(x - y) - 21 {(x - y)}^{2}$

Let assume that

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: x - y = z \: \: }}}$

So, above expression can be rewritten as

$\rm \:  =  \: 30 - 17z - 21 {z}^{2}$

$\rm \:  =  \: - ( {21z}^{2} + 17z - 30)$

Niw, we use the concept of Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to find numbers p and q such that p + q = b and pq = ac.

After finding p and q, we split the middle term in the quadratic expression as px + qx and get the required factors after regrouping the terms.

So, here we have to find p and q in such a way that

$\rm :\longmapsto\:p + q = 17$

and

$\rm :\longmapsto\:pq = - 630$

So, 630 can be rewritten in factor form as

$\rm :\longmapsto\:pq = -7 \times 3 \times 3 \times 2 \times 5$

$\rm :\longmapsto\:pq = -(7 \times 5) \times( 3 \times 3 \times 2)$

$\rm :\longmapsto\:pq = -(35) \times(18)$

So,

$\rm\implies \:p = 35 \: \: and \: \: q = - 18$

So, given expression can be rewritten as

$\rm \:  =  \: - (21 {z}^{2} + 35z - 18z - 30)$

$\rm \:  =  \: - [7z(3z + 5) - 6(3z + 5)]$

$\rm \:  =  \: - [(3z + 5)(7z - 6)]$

Now substituting the value of z = x + y, we get

$\rm \:  =  \: - [3(x + y) + 5][7(x + y) - 6]$

$\rm \:  =  \: - [3x + 3y + 5][7x + 7y - 6]$

Hence,

$\bf :\longmapsto\:30 - 17(x - y) - 30 {(x - y)}^{2} \\ \\ \bf \:  =  \: - [3x + 3y + 5][7x + 7y - 6] \: \: \: \: \:$

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)