Question

please someone solve this ​

Answer 1

Answer:

$= > {x}^{2} - 23x + 132 \\ \\ = > {x}^{2} - 12 x - 11x + 132 \\ \\ = > x(x - 12) - 11(x - 12) \\ \\ = > (x - 11)(x - 12) \\ \\ (x - 12)(x - 11) = 0 \\ \\ x = 12 \\ x = 11 \: \: ans...$

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Answer 2

Question :-

Factorise ;

x² - 23x + 132

Answer :-

Given :-

Quadratic polynomial : x² - 23x + 132

we need to factorise the above polynomial

So,

The method which we need to use is Splitting the middle term

Because , the constant term is not a perfect square .

If the constant term is a perfect square , then we can use identities to solve it !

Here,

This the procedure on how to factorise the above expression .

$\rule{200}{2}$

$\rightarrowtail{\tt{ {x}^{2} - 23x + 132 }}$

Here,

We need to split the middle term in such a way that the sum of the them should be equal to the middle term and their product should be equal to the constant term

So,

23 can be split into - 12 & - 11

Because, - 12x - 11x = - 23x & - 12 x - 11 = 132

( here we need to attach the variable depending upon the variable mentioned in the question )

Hence,

$\rightarrowtail{\tt{ {x}^{2} - 12x - 11x + 132 }}$

$\rightarrowtail{\tt{ x ( x - 12 ) - 11 ( x - 12 ) }}$

$\implies{\tt{ (x - 12 ) ( x - 11 ) }}$

Therefore,

The given quadratic expression can be factorised into ( x - 12 ) & ( x - 11 )

Actually these are the factors of the given quadratic polynomial .

Verification :-

Now, let's check whether our factorisation is correct or wrong .

For that ; just multiply the both factors and the result should be the given above quadratic polynomial

So,

( x - 12 ) x ( x - 11 )

Using the identity ;

( x + a ) ( x + b ) = x² + x ( a + b ) + ab

So,

( x )² + x ( - 12 - 11 ) + (- 12)(- 11 )

x² + x ( - 23 ) + 132

x² - 23x + 132

$\large{\underline{\text{Hence verified }}}{\huge{\checkmark}}$