SOLUTION: I need help with the following:
1. t^4 - 19t^2 + 48
I got 4,-4, sqrt{3}, negative sqrt{3} for the zeros using synthetic but it was wrong.
When I factored the equation
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-> SOLUTION: I need help with the following:
1. t^4 - 19t^2 + 48
I got 4,-4, sqrt{3}, negative sqrt{3} for the zeros using synthetic but it was wrong.
When I factored the equation
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1. t^4 - 19t^2 + 48
I got 4,-4, sqrt{3}, negative sqrt{3} for the zeros using synthetic but it was wrong.
When I factored the equation out completely i got (x+4)(x-4)(x+3)(x-3) but it was wrong.
2. f(x) = x^3 + 6x^2 + 11x + 12. Write the polynomial as the product of linear factors.
#2 will have negative roots, and maybe some positive roots. Check using synthetic division for pos & neg of 1, 2, 3, 4, 6.
The only REAL root is for the linear binomial factor, . ( I used Google Search Engine's graphing of function feature instead of actually doing synthetic division). "product of linear binomials"? No. Not unless you want two complex factors, too. You can use general solution for quadratic equation for that.
Synthetic Division checking root gives you coefficients
1, 2, 3
Meaning the factorization .
Roots for the quadratic factor are and . General solution for quadratic formula will give those.
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