factorise
![x2 - x - 110 x2 - x - 110](https://tex.z-dn.net/?f=x2+-+x+-+110+)
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Answered by
5
Answer :
Now, x² - x - 110
= x² - (11 - 10)x - 110, since 110 = 11 × 10
= x² - 11x + 10x - 110
= x (x - 11) + 10 (x - 11)
= (x - 11) (x + 10),
which is the required factorization.
#MarkAsBrainliest
Now, x² - x - 110
= x² - (11 - 10)x - 110, since 110 = 11 × 10
= x² - 11x + 10x - 110
= x (x - 11) + 10 (x - 11)
= (x - 11) (x + 10),
which is the required factorization.
#MarkAsBrainliest
SilentToper:
thanxx bro
Answered by
1
Find numbers whose sum is -1 and product is -110.
They are -11 & 10.
![{x}^{2} - x - 110 = {x}^{2} - 11x + 10x - 110 \\ x(x - 11) + 10(x - 11) \\ (x - 11)(x + 10) \: are \: \: the \: \: factors {x}^{2} - x - 110 = {x}^{2} - 11x + 10x - 110 \\ x(x - 11) + 10(x - 11) \\ (x - 11)(x + 10) \: are \: \: the \: \: factors](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B2%7D++-+x+-+110+%3D++%7Bx%7D%5E%7B2%7D++-+11x+%2B+10x+-+110+%5C%5C+x%28x+-+11%29+%2B+10%28x+-+11%29+%5C%5C+%28x+-+11%29%28x+%2B+10%29+%5C%3A+are+%5C%3A++%5C%3A+the+%5C%3A++%5C%3A+factors)
They are -11 & 10.
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