Question

Divide 144x⁴y³(x²-7x+12) by -54x²y(x-4)​

Answer 1

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

$\: \: \: \: \: \: \: \: \bull \sf \: \dfrac{144 {x}^{4} {y}^{3} ( {x}^{2} - 7x + 12) }{ - 54 {x}^{2} y(x - 4)}$

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

↝ We first factorize the numerator by using the concept of splitting of middle terms.

↝ Splitting of middle terms :-

• In order to factorize  x² + bx + c we have to find numbers p and q such that p + q = b and pq = c.

• After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

Now,

↝ Consider,

$\rm :\longmapsto\: {x}^{2} - 7x + 12$

↝ Now, we have to find p and q in such a way that pq = 12 and p + q = - 7.

↝ So, p and q can take the values - 3 and - 4.

Therefore,

$\rm :\longmapsto\: {x}^{2} - 7x + 12 = {x}^{2} - 3x - 4x + 12$

$\rm :\longmapsto\: {x}^{2} - 7x + 12 =x(x - 3) - 4(x - 3)$

$\rm :\longmapsto\: {x}^{2} - 7x + 12 = \: (x - 3)(x - 4)$

Now,

$\rm :\longmapsto\:144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = {2}^{4} \times {3}^{2}$

$\rm :\longmapsto\:54 = 2 \times 3 \times 3 \times 3 = 2 \times {3}^{3}$

Now,

↝Consider,

$\rm :\longmapsto\:\sf \: \dfrac{144 {x}^{4} {y}^{3} ( {x}^{2} - 7x + 12) }{ - 54 {x}^{2} y(x - 4)}$

$\sf \: = \sf \: \dfrac{ {2}^{4} \times {3}^{2} \times {x}^{4} \times {y}^{3} (x - 3)(x - 4) }{ - 2 \times {3}^{3} \times {x}^{2} \times y \times (x - 4)}$

↝ Now, on cancelation of like terms, we get

$\sf \: = - \: \dfrac{ {2}^{3} \times {x}^{2} \times {y}^{2} \times (x - 3) }{3}$

$\sf \: = - \: \dfrac{8}{3} \: {x}^{2} \: {y}^{2} \: (x - 3)$

$\overbrace{ \underline { \boxed { \rm \therefore \: \sf \: \dfrac{144 {x}^{4} {y}^{3} ( {x}^{2} - 7x + 12) }{ - 54 {x}^{2} y(x - 4)} = - \: \dfrac{8}{3} \: {x}^{2} \: {y}^{2} \: (x - 3)}}}$

Additional Information :-

$\underline{ \boxed{ \bf {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}$

$\underline{ \boxed{ \bf {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}$

$\underline{ \boxed{ \bf {x}^{2} - {y}^{2} = (x + y)(x - y)}}$