what i did wrong ........the picture is answer ...and question is 1. factorise x^3-3x^2-9x-5
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x³ - 3x² - 9x - 5
=> x³ + x² - 4x² - 4x - 5x - 5
[ -3x² = x² - 4x² and -9x = -4x - 5x ]
=> x²(x + 1 ) - 4x( x + 1 ) - 5(x + 1 )
=> ( x + 1 ) ( x² - 4x - 5 )
=> ( x + 1 ) [ x² - ( 5 - 1 ) x - 5 ]
=> ( x + 1 ) (x² - 5x + x - 5 )
=> ( x + 1 ) [ x( x - 5 ) + ( x - 5 )]
=> ( x + 1 ) ( x + 1 ) ( x - 5)
They are factors, Match your solution.
=> x³ + x² - 4x² - 4x - 5x - 5
[ -3x² = x² - 4x² and -9x = -4x - 5x ]
=> x²(x + 1 ) - 4x( x + 1 ) - 5(x + 1 )
=> ( x + 1 ) ( x² - 4x - 5 )
=> ( x + 1 ) [ x² - ( 5 - 1 ) x - 5 ]
=> ( x + 1 ) (x² - 5x + x - 5 )
=> ( x + 1 ) [ x( x - 5 ) + ( x - 5 )]
=> ( x + 1 ) ( x + 1 ) ( x - 5)
They are factors, Match your solution.
Answered by
6
ʜᴇʟʟᴏ ᴍᴀᴛᴇ!
Let's take 5 as value of x, as you did
p(x) = (5)³ - 3(5)² - 9(5) - 5
0 = 125 - 75 - 45 - 5
0 = 0
So, ( x - 5 ) is factor of p(x)
Now, Divide the p(x) by ( x - 5 ),
Here, I have attached an image. In attatchment you will get full solution.
Answer is ( x - 5 )( x + 1 )( x + 1 )
What mistake you did?
=> You left the answer at x - 5 which is wrong. It has further solution to considered valid.
Hope it helps
Let's take 5 as value of x, as you did
p(x) = (5)³ - 3(5)² - 9(5) - 5
0 = 125 - 75 - 45 - 5
0 = 0
So, ( x - 5 ) is factor of p(x)
Now, Divide the p(x) by ( x - 5 ),
Here, I have attached an image. In attatchment you will get full solution.
Answer is ( x - 5 )( x + 1 )( x + 1 )
What mistake you did?
=> You left the answer at x - 5 which is wrong. It has further solution to considered valid.
Hope it helps
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abhi569:
Actually, teacher marked it as wrong because question is 'Factorise x³-3x²-9x -5'
Not to tell whether (x - 5) is a factor or not
yeah ...kk
Hm
But... I think... You have seen the answer from last of the book
That's why, you directly wrote ( x - 5)
nope
Hm
its test ....hw can i see frm textbook
It is not my problem.... May you had seen from book already
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