Factorise 9 - 2√3x + x²
Answers
Answer:
Theory
If your binomial equation is in the form:
Then you can find the factor only when there exists some m and n such that
a x c = m x n and m+n = b.
If any such combination exists, you factor the equation in a simpler form.
Example
Example : 3x² -16x -12
Here a = 3, b = -16 and c = -12
Now there is one combination of m and n which completes the described condition. And the combination is m = 2 and n = -18.
3 x -12 = 2 x -18 which satisfies the condition {a x c = m x n}
2 + (-18) = -16 which satisfies the condition {m + n = b }.
Now if such condition exists, we should write it in easy format where we expand b in terms of m and n which implies:
==> 3x² -16x -12 = 3x² + 2x -18x -12
==> = x(3x + 2) -6(3x + 2)
==> = (3x + 2)(x - 6) -------- Solved
Numerator factoring
The same method can be implemented in solving the numerator of the given equation which is 9 - 2√3x - x². Or we can write it in the form of
: == > -x² -2√3x + 9
a = -1
b = -2√3
c = 9
a x c = (-1) x 9 = -9
while b = -2√3. Well i can see a combination which satisfies the factorization condition and for which m and n will be:
m = -3√3
n = √3
m x n = (-3√3) x (√3) = -3 x 3 = -9 = a x c {First condition satisfied}
m + n = (-3√3) + (√3) = -2√3 = b {Second condition satisfied}
Now if such condition exists, we should write it in easy format where we expand b in terms of m and n which implies:
==> -x² -2√3x + 9 = -x² -3√3x + √3x + 9
==> = -x(x + 3√3) + √3(x + 3√3)
==> = (x + 3√3) (-x + √3) {numerator solved}
Denominator Factoring
The equation in denominator is 3 - x². This can be suited to 2 formulas
a² - b² = (a + b) (a - b)
The same ax² +bx + c (I leave factoring 3 - x² with the given formula to you as a home work and going with formula 1)
Here, if we express 3 - x² in a² - b² form, it will look like:
(√3)² - x² which implies that
a = √3 and b = x. Hence :
3 - x² = (√3)² - x² = (√3 + x)(√3 - x) {Denominator Solved }
Answer:
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9-2√3x+x²
