X2-2√5x+3 factorize it by spilitting the middle term
Answers
You do it the usual way.
The coefficient of the middle term is negative, but the constant term is positive. Hence, you must look for two numbers whose sum is 25–√ and product is 3.
The product being rational, there must be some irrational conjugate multiplication happening here. If two irrational conjugates add to 25–√, they must be 5–√±c, where c is a number we have to determine.
Their product is 3. Therefore, (5–√)2−c2=3. Hence, the two numbers required are 5–√±2–√.
Express the middle term and the constant in terms of these numbers.
x2−25–√x+3=x2−(5–√+2–√+5–√−2–√)x+(5–√+2–√)(5–√−2–√)
⟹x2−25–√x+3=x2−(5–√+2–√)x−(5–√−2–√)x+(5–√+2–√)(5–√−2–√)
⟹x2−25–√x+3=x[x−(5–√+2–√)]−(5–√−2–√)[x−(5–√+2–√)]
⟹x2−25–√x+3=[x−(5–√+2–√)][x−(5–√−2–√)]
Alternatively (and this is at worst), you could simply use the quadratic formula to determine its zeros, and then express the polynomial as a product of two linear factors.
Alternatively (and this is at worst), you could simply use the quadratic formula to determine its zeros, and then express the polynomial as a product of two linear factors.
−(−25–√)±(−25–√)2−4×1×3−−−−−−−−−−−−−−−−√2×1=5–√±2–√
⟹x2−25–√x+3=[x−(5–√+2–√)][x−(5–√−2–√)]