Question

please answer
this .I hav ebeen trying for long

Answer 1

Question:

Simplify

• $(a)\large{ \frac{3 {x}^{3} - 6 {x}^{2} - 24x }{(x - 4)(x + 2)}}$

• $(b) \large{\frac{ {p}^{4} - 256 }{ {p}^{3} + 4 {p}^{2} + 16p + 6y}}$

Answer:

$\large\bold\red{(a)\:3x}\\\\\large\bold\red{(b)\frac{(p + 4)(p - 4)( {p}^{2} + 16)}{p( {p}^{2} + 4p + 16) + 6y }}$

Step-by-step explanation:

$(a ) \frac{3 {x}^{3} - 6 {x}^{2} - 24x }{(x - 4)(x + 2)} \\ \\ = \frac{3x( {x}^{2} - 2x - 8)}{(x - 4)(x + 2)} \\ \\ = \frac{3x( {x}^{2} - 4x + 2x - 8) }{(x - 4)(x + 2)} \\ \\ = \frac{3x(x(x - 4) + 2(x - 4))}{(x - 4)( x + 2)} \\ \\ = \frac{3x(x - 4)(x + 2)}{(x - 4)(x + 2)} \\ \\ = \large \boxed{ \bold{3x}}$

$(b) \frac{ {p}^{4} - 256}{ {p}^{3} + 4 {p}^{2} + 16p + 6y } \\ \\ = \frac{ {( {p}^{2} )}^{2} - {(16)}^{2} }{ p({p}^{2} + 4p + 16) + 6y} \\ \\ = \frac{( {p}^{2} + 16)( {p}^{2} - 16) }{p( {p}^{2} + 4p + 16) + 6y } \\ \\ = \frac { ( {p}^{2} + 16)( {p}^{2} - {4}^{2} )}{p( {p}^{2} + 4p + 16) + 6y } \\ \\ = \large \boxed{ \bold{ \frac{(p + 4)(p - 4)( {p}^{2} + 16)}{p( {p}^{2} + 4p + 16) + 6y } }}$

Answer 2

Question :---- Above ↑↑↑↑

Solution of Question (1)

$\frac{3 {x}^{3} - 6 {x}^{2} - 24x }{(x - 4)(x + 2)} \\ \\ taking \: 3x \: common \: from \: denominator \\ we \: get \\ \\ \implies \: \frac{3x( {x}^{2} - 2x - 8)}{(x - 4)(x + 2)} \\ \\ now \: solving \: ( {x}^{2} - 2x - 8) \: first \\ by \: splitting \: the \: middle \: term \\ \\ \implies \: {x}^{2} - 4x + 2x - 8 \\ \\ \implies x(x - 4) + 2(x - 4) \\ \\ \implies \: (x - 4)(x + 2) \\ \\ putting \: in \: numerator \: now \\ \\ \implies \: \frac{3x \cancel{(x - 4)(x + 2)}}{\cancel{(x - 4)(x + 2)}} \: \\ \\ \implies \: \red{\bold{3x}}$

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Formula used here :---

• (a⁴ - b⁴) = (a²+b)²(a-b)(a+b)

Lets solve numerator and denominator separately ,,

p⁴ - 256

→ p⁴ - (4)⁴

→ (p²+4²)(p-4)(p+4)

→ (p²+16)(p-4)(p+4)

Now, denominator ,,,

p³ + 4p² +16p +6y

taking p common ,

→ p(p² + 4p + 16) + 6y

so, we get,,,

$\red{\bold{ \frac{( {p}^{2} + 16)(p - 4)(p + 4) }{p(p^{2} + 4p + 16) + 6y } }}$

(Hope it Helps you)