Asked by shelly
I need help in understanding these operations factor by grouping 8x^3-56^2-5x+35
Answers
Answered by
drwls
Is the second term supposed to be 56x^2? If so, then use the following as an example to factor by grouping
x^3 - 3x^2 + 2x - 6
= (x^3 - 3x^2) + (2x - 6). Now factor out the common factors from each group such as x^2(x - 3) + 2(x - 3). That can be rewritten as (x-3)*(x^2 + 2)
In your case,
8x^3-56^2-5x+35 = 8x^2(x-7)-5(x-7)
= (8x^2-5)(x-7)
The roots are x=7 and x = +/-sqrt(5/8)
x^3 - 3x^2 + 2x - 6
= (x^3 - 3x^2) + (2x - 6). Now factor out the common factors from each group such as x^2(x - 3) + 2(x - 3). That can be rewritten as (x-3)*(x^2 + 2)
In your case,
8x^3-56^2-5x+35 = 8x^2(x-7)-5(x-7)
= (8x^2-5)(x-7)
The roots are x=7 and x = +/-sqrt(5/8)
Answered by
drwls
This cubic equation was "set up" to work. In by far most cases, with random integer coefficients, it doesn't.
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