x2 – 3x + 2 এবং x2 – 5x + 6 এর লসাগু = কত?
Solution
Correct Answer: Option A
১ম রাশি = x² -3x+2
= x²-2x-x+2
= x(x-2)-1(x-2)
= (x-2)(x-1)
২য় রাশি = x² -5x+6
= x²-3x-2x+6
= x(x-3)-2(x-3)
= (x-3)(x-2)
∴ নির্ণেয় ল .সা .গু = (x-1) (x-2) (x-3)
= x²-2x-x+2
= x(x-2)-1(x-2)
= (x-2)(x-1)
২য় রাশি = x² -5x+6
= x²-3x-2x+6
= x(x-3)-2(x-3)
= (x-3)(x-2)
∴ নির্ণেয় ল .সা .গু = (x-1) (x-2) (x-3)
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