Question

3(x^2-5x)^2-2(x^2-5x+5)-6 find factorize polynomial​

Answer 1

Answer:

Required factorised polynomial is ( 3x^2 - 15x - 8 )( x^2 - 5x + 2 ) .

Step-by-step-explanation:

Given,

3( x^2-5x )^2 - 2( x^2-5x+5 ) - 6

Let,

x^2 - 5x = a

Now, { continued }

= > 3( a )^2 - 2( a + 5 ) - 6

= > 3a^2 - 2a - 10 - 6

= 3a^2 - 2a - 16

= > 3a^2 - ( 8 - 6 )a - 16

= > 3a^2 - 8a + 6a - 16 = 0

= > a( 3a - 8 ) + 2( 3a - 8 )

= > ( 3a - 8 )( a + 2 )

If it is not equal to 0 : ( it is a polynomial )

= > { 3( x^2 - 5x } - 8 } ( x^2 - 5x + 2 )

= > ( 3x^2 - 15x - 8 )( x^2 - 5x + 2 )

If it is equal to 0 :

= > a = 8 / 3 or - 2

Substituting the value of a :

= > x^2 - 5x = 8 / 3 Or x^2 - 5x = - 2

= > 3x^2 - 15x = 8 Or x^2 - 5x + 2 = 0

= > 3x^2 - 15x - 8 = 0 Or x^2 - 5x + 2 = 0

Using quadratic equation : If an equation is written as ax^2 + bx + c = 0 then possible values of x can be determined by using x = { - b ± √( b^2 - 4ac ) } / 2a

On the basis of the formula given above :

Case 1 : 3x^2 - 15x - 8 = 0

= > x = { 15 ± √{ 225 + 4( 8 × 3 ) } } / 2( 3 )

= > x = { 15 ± √321 } / 6

Similarly,

Case 2 : If x^2 - 5x + 2 = 0

= > x = { 5 ± √{ 25 - 4( 2 ) } / 2

= > x = ( 5 ± √17 ) / 2

Hence the required values of x are ( 5 ± √17 ) / 2 and ( 15 ± √321 ) / 6.

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Given,

3(x² - 5x)² - 2(x² - 5x + 5) - 6

Assume ,

x² - 5x = a

Also,

= 3(a)² - 2(a + 5) - 6

= 3a² - 2a - 10 - 6

= 3a² - 2a - 16

= 3a² - (8 - 6)a - 16

3a² - 8a + 6a - 16 = 0

= a(3a - 8) + 2(3a - 8)

= (3a - 8)(a + 2)

3a - 8 = 0

3a = 8

$\tt{\rightarrow a=\dfrac{8}{3}}$

Also,

a + 2 = 0

a = -2

If it is not equal to 0 :-

Then,

= [3(x² - 5x) - 8] (x² - 5x + 2)

= (3x² - 15x - 8)(x² - 5x + 2)

If it is equal to 0 :-

$\tt{\rightarrow a=\dfrac{8}{3}\;,\;-2}$

Putting the value of a :-

$\tt{\rightarrow x^2-5x=\dfrac{8}{3}}$

x² - 5x = -2

3x² - 15x = 8

x² - 5x + 2 = 0

3x² - 15x - 8 = 0

x² - 5x + 2 = 0

Quadratic Equation :

$\tt{\rightarrow\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}$

First :-

3x² - 15x - 8 = 0

$\tt{\rightarrow x=\dfrac{15\pm \sqrt{225+4\times 8\times 3}}{2\times 3}}$

$\tt{\rightarrow x=\dfrac{15\pm \sqrt{321}}{6}}$

Also

Second :-

If x² - 5x + 2 = 0

$\tt{\rightarrow x=\dfrac{5\pm \sqrt{25-4(2)}}{2}}$

$\tt{\rightarrow x=\dfrac{5\pm \sqrt{17}}{2}}$