Where does 1/4 come from in this linear factorization?
- Thread starter Addie1103
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Using the zeroes, we may write:
[MATH]f(x)=(x-1)\left(x+\frac{1}{2}-\frac{\sqrt{11}i}{2}\right)\left(x+\frac{1}{2}+\frac{\sqrt{11}i}{2}\right)[/MATH]
Next, we can factor 1/2 from the two rightmost factors:
[MATH]f(x)=(x-1)\frac{1}{2}\left(2x+1-\sqrt{11}i\right)\frac{1}{2}\left(2x+1+\sqrt{11}i\right)[/MATH]
The commutative property of multiplication allows us to bring those two 1/2's up front:
[MATH]f(x)=\frac{1}{2}\cdot\frac{1}{2}(x-1)\left(2x+1-\sqrt{11}i\right)\left(2x+1+\sqrt{11}i\right)[/MATH]
And finally, we may multiply them together to get:
[MATH]f(x)=\frac{1}{4}(x-1)\left(2x+1-\sqrt{11}i\right)\left(2x+1+\sqrt{11}i\right)[/MATH]
[MATH]f(x)=(x-1)\left(x+\frac{1}{2}-\frac{\sqrt{11}i}{2}\right)\left(x+\frac{1}{2}+\frac{\sqrt{11}i}{2}\right)[/MATH]
Next, we can factor 1/2 from the two rightmost factors:
[MATH]f(x)=(x-1)\frac{1}{2}\left(2x+1-\sqrt{11}i\right)\frac{1}{2}\left(2x+1+\sqrt{11}i\right)[/MATH]
The commutative property of multiplication allows us to bring those two 1/2's up front:
[MATH]f(x)=\frac{1}{2}\cdot\frac{1}{2}(x-1)\left(2x+1-\sqrt{11}i\right)\left(2x+1+\sqrt{11}i\right)[/MATH]
And finally, we may multiply them together to get:
[MATH]f(x)=\frac{1}{4}(x-1)\left(2x+1-\sqrt{11}i\right)\left(2x+1+\sqrt{11}i\right)[/MATH]
Harry_the_cat
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The zero of 1 gives the factor (x - 1).
The zero of \(\displaystyle \frac{-1}{2}+i \frac {\sqrt11}{2} \) gives the factor \(\displaystyle (x + \frac{1}{2} - i \frac{\sqrt 11}{2}) = \frac{1}{2}(2x + 1 - i\sqrt{11})\) when \(\displaystyle \frac{1}{2}\) is factorised out
Same with the other complex root.
The zero of \(\displaystyle \frac{-1}{2}+i \frac {\sqrt11}{2} \) gives the factor \(\displaystyle (x + \frac{1}{2} - i \frac{\sqrt 11}{2}) = \frac{1}{2}(2x + 1 - i\sqrt{11})\) when \(\displaystyle \frac{1}{2}\) is factorised out
Same with the other complex root.
