9 + 15mfactorisation of algebraic expression that contain common factor
2 answers:
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Step-by-step explanation:
9514 1404 393
Answer:
- real: -1, 2; complex: +i, -i
- 1, 3, 4
Step-by-step explanation:
1. The graph (red) shows the only real zeros to be -1 and 2. When the corresponding factors are divided from the function, the remaining factor is the quadratic (x^2 +1), which has only complex roots. The quadratic is graphed in green.
The linear factorization is ...
f(x) = (x +1)(x -2)(x -i)(x +i)
The roots are -1, 2, -i, +i.
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2. The graph (blue) shows the zeros are 1, 3, 4.
You observe that the sum of coefficients is zero, so x=1 is a root. Factoring that out gives the quadratic (x^2 -7x +12), which you recognize factors as
(x -3)(x -4) . . . zeros of 3 and 4
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I have attached a spreadsheet that does synthetic division. There are web sites that will do this, too. The tables shown correspond to f1(x)/(x-2) and f2(x)/(x-1). When you fill in the zero and coefficients, the built-in formulas do the rest.
