Question

Factorise xsquare+ 5√3x + 12

Answer 1

$\sf answer=\boxed{\bf\large\orange{\underline{(x + \sqrt{3}) (x+ 4\sqrt{3})}}}$

GIVEN:-

$\sf \bf {x^2 + 5\sqrt{3x}+ 12}$

TO FIND :-

↦ factors

SOLUTION :-

$\sf \bf {x^2 + 5\sqrt{3x}+ 12}$

AS WE KNOW,

$\boxed{\sf\underline{(x + a)(x+b)=x^{2} + (a+b)x+ab}}$

$\sf\bold {x^2 + \sqrt{3x} + 4\sqrt{3x} + 12x^2}$

$\sf\bf x (x + \sqrt{3}) + 4\sqrt{3} ( x + \sqrt{3} )$

$\boxed{\bf\large\underline\pink{{(x + \sqrt{3}) (x+ 4\sqrt{3})}}}$

LET'S EXPLORE MORE

$\bf\red{{(a+b)} ^{2} ⟹{a}^{2}+2ab+{b}^{2}}$

$\bf\green{(a - b)^{2} \implies \: {a}^{2} - 2ab + {b}^{2} }$

$\bf\red{(a + b)(a - b) \implies \: {a}^{2} - {b}^{2} }$

$\bf\green{(x + a)(x \: + b) \implies \: {x}^{2} + (a + b)x + ab}$

$\bf\red{(a + b)^{3} \implies \: {a}^{3} + 3 {a}^{2} b +3a {b}^{2} + {b}^{3} }$

$\bf\green{(a - b)^{3} \implies \: {a}^{3} - 3 {a}^{2} b +3a {b}^{2} - {b}^{3}}$

Answer 2

$\red{\large{\underline{\underline{ \rm{Question }}}}}$

Factorise :-

${x}^{2} + 5 \sqrt{3x} + 12$

How to Solve :-

When a trinomial is of the form ax² + bx + c or ( a + bx + cx² ), split b (the coefficient of x in the middle term) into two parts such that the sum of these two parts is equal to b and the product of these two parts is equal to the product of a and c. Then factorise by the grouping method.

$\red{\large{\underline{\underline{ \rm{</strong><strong>Solu</strong><strong>tion</strong><strong> }}}}}$

The given equation is

${x}^{2} + 5\sqrt{3x} + 12$

We split $5\sqrt{3}$ into two parts such that the sum of these two parts is $5\sqrt{3}$ and the product is $12$.

As :

$(4\sqrt{3} + \sqrt{3} ) = 5\sqrt{3}$

$(4 \sqrt{3} \times \sqrt{3} ) = 12$

We have,

${x}^{2} + 5\sqrt{3x} + 12$

$= {x}^{2} + 4 \sqrt{3x} + \sqrt{3x} + 12$

$= x(x + 4\sqrt{3} ) \: \sqrt{3} (x + 4\sqrt{3} )$

$= { \green{ \underline{\boxed{(x + 4\sqrt{3} ) \: (x + \sqrt{3}) }}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\rm{ \underline{ \underline{</u><u>Some</u><u> </u><u>\</u><u>:</u><u> </u><u>I</u><u>dentities</u><u>:</u><u>}}}$

$\purple{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}$

$\blue{ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}$

$\purple{ (a+ b) \: (a - b) = {a}^{2} - {b}^{2} }$

$\blue{(x + a) \: (x + b) = {x}^{2} + (a + b)x + ab}$

$\purple{ {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} ) }$

$\blue{ {(a + b)}^{2} - {(a - b)}^{2} = 4ab}$