Math comnic project
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- Hi ms can u please explain how to factor by grouping 6n^3+3n^2+8n+4, i have been trying to solve it for the past 3 houra
- ok the first step were gonna take the greatest common factor so were gonna have (6n^3+3n^2)(+8n+4) then its gonna be (6n^3+3n^2)4(2n+4) as the common factor then were gonna do the same too (6n^3+3n^2) so its gonna be 3n^2(2n+1)4(2n+1) now we have two of (2n+1) but we also have two digits of the common factors so a easy way to do this is we gonna combine the common factors togther and since we have two (2n+1) were gonna only use one so the final answer is gonna be (3n^2+4)(2n+1)
- Ms can u please explain how can i factor the difference between 2 perfect squared x^2-25
- When an expression can be viewed as the difference of two perfect squares, i.e. a²-b², then we can factor it as (a+b)(a-b). For example, x²-25 can be factored as (x+5)(x-5). This method is based on the pattern (a+b)(a-b)=a²-b², which can be verified by expanding the parentheses in (a+b)(a-b).
- Thank you ms i understand it so much betterrrr
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