Question

ques5) factorise:- x^3-2x^2-c+2​

Answer 1

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Valid \: QuestioN...}}}}}$

⋆ Factorise the following expression:

• ${\sf{x^3-2x^2-x+2}}$

${\bigstar \:{\pmb{\sf{\underline{GivEn \: Expression...}}}}}$

• ${\sf{x^3-2x^2-x+2}}$

${\bigstar \:{\pmb{\sf{\underline{What \: to \: do...}}}}}$

• Have to factorise

${\bigstar \:{\pmb{\sf{\underline{UsiNg \: Concepts...}}}}}$

⋆ Middle term splitting method

⋆ Hit and trial method

⋆ Remainder theorm

${\bigstar \:{\pmb{\sf{\underline{Something \: about \: ConcEpts...}}}}}$

⋆ Middle term splitting method:

• Let us learn something about middle term splitting method by taking an example.

• Let the given expression is ${\pmb{\sf{\red{x^{2}+13x+40}}}}$ We have to solve this expression by using middle term splitting method. Let us solve this example.

${\bigstar \:{\pmb{\sf{\purple{\underline{\underline{Middle \: term \: splitting \: method...}}}}}}}$

${\sf{:\implies x^{2}+13x+40}}$

${\sf{:\implies x^{2} + (8+5)x + 40}}$

${\sf{:\implies x^{2} + 8x + 5x + 40}}$

${\sf{:\implies x(x) + (2 \times 2 \times 2) + 5(x) + (2 \times 2 \times 2)}}$

${\sf{:\implies x(x) + (4 \times 2) + 5(x) + (4 \times 2)}}$

${\sf{:\implies x(x) + (8) + 5(x) + (8)}}$

${\sf{:\implies x(x+8) + 5(x+8)}}$

${\sf{:\implies (x+8) (x+5) = 0}}$

${\sf{:\implies x \: = 0-8 \: or \: x =\: 0-5}}$

${\sf{:\implies x = \: -8 \: or \: x \: = -5}}$

${\pmb{\tt{Henceforth, \: x \: = -8 \: or \: -5}}}$

${\pmb{\tt{Henceforth, \: solved!}}}$

Will you understand anything regards this method? 1% , 2% or else, anything? (Anyway) Very good try it is! Now let me explain this to you in simple way!

• What is given expression? x²+13x+40 What is the middle term in this expression? It's 13x. Now let's see about the 1st and 3rd term. What are they? x² and 40. (Nice!) Now what is the product of x² and 40? It's 40x² Remember that we have to split the middle term in that way that the two contain that we obtain by splitting given us the product of 1st and 3rd term as here the product is 40x²!

• And also those two numbers when subtracted or added will give the middle term! Ok firstly, how can we get 40x² as product by splitting 13x? Firstly what are factor of 40? ±1 , ±2 , ±4 , ±5 , ±8 , ±10 , ±20 There are the factors of 40. Now let's see what are the two numbers that we multiply and get 40. There are 8 and 5 as 8 × 5. Now let's check that are they given 13 while subtracting or adding. 8 - 5 = 3(Error) , 8 + 5 = 13, 13 = 13. In short we get the two numbers that give the process correctly! But don't forget that the product….expression is 40x² so terms be (8+5)x Afterwards as you already know about regrouping method, so then have to use it and at last of using correctly then we will get our final result!

⋆ Hit and trial method:

• Remember trial and error method? Nice! It is just same like that but here is the concept of polynomial so we have to check he zeroes by this method!

⋆ Remainder theorm:

• Remainder theorm is important! I will tell you about this while solving question.

${\bigstar \:{\pmb{\sf{\underline{SolutioN...}}}}}$

~ Firstly, are you confused that why we have to use this type of so much methods, as we can use it by using just middle term splitting method? We have to use so much methods here because see carefully that what is expression given, it is a cubic polynomial! Means it's factors are also three not as quadratic polynomial!

~ To solve this firstly we have to check that in the given polynomial that is ${\sf{x^3-2x^2-x+2}}$ what is the constant term l,the constant term is 2 here. So what are factors of 2? There are ±1 and ±2. Now using its factors & hit and trial method, we get the following results!

• As 1 is one of the factor of 2 so let's use it!

⟹ x+1 = 0

⟹ x = 0-1

⟹ x = -1

• Now putting -1 in polynomial at the place of x we get,

⟹ (-1)³ - 2(-1)² - (-1) + 2

⟹ -1 -2(1) -(-1) + 2

⟹ -1-2+1+2

⟹ -3 + 3

⟹ 0

• If we are getting 0 at last then we get our first factor of the given polynomial. Henceforth, (x-1) is one of the factor of given polynomial.

~ Now dividing given polynomial by obtained factor we get,

⟹ Kindly refer the attachment 1st for see the methods of division. And also refer to attachment 2nd, it is regard working note.

~ Now as get quotient as x²-x-2, we have to factorise it by using middle term splitting method and afterwards we get two more factors of the polynomial.

⟹ x²-x-2

⟹ x²-2x+1-2

⟹ x(x-2) + 1(x-2)

⟹ (x-2) (x+1)

• The two more factors are (x-2) (x+1) of the given polynomial.

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Solution:

Factor's = (x-1) (x-2) (x+1)