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horrorfan [7]
3 years ago
14

Help please #12 Solve using factoring

Mathematics
2 answers:
Serga [27]3 years ago
5 0

(R-2)x(r+2) that what I think it is

zimovet [89]3 years ago
5 0

You have to use the sum/difference formula

(a+b)(a-b)=a^2-b^2

In your case, r^2-4 is a difference of squares, so it can be factored with the sum and difference of the roots:

r^2-4 = (r+2)(r-2)

So, we have

r^2-4 = 0 \iff (r+2)(r-2) = 0

A multiplication is zero if and only if one of the factors is zero, so we have

(r+2)(r-2) = 0 \iff r+2=0\ \lor\ r-2=0 \iff r=-2\ \lor\ r=2

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What are the restrictions for a? 2a^2+a-15/5a^2+16a+3
koban [17]

ANSWER


The restrictions are

a\ne -3,a\ne -\frac{1}{5}


EXPLANATION


We were given the rational function,


\frac{2a^2+a-15}{5a^2+16a+3}


The function is defined for all values of a, except



5a^2+16a+3=0


This has become a quadratic trinomial, so we need to split the middle term.


We do that by multiplying the coefficient of x^2 which is 5 by the constant term which is 3. This gives us 15.


The factors of 15 that adds up to 16 are 1 and 15.


We use these factors to split the middle term.




5a^2+15a+a+3=0


We now factor to get,


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We factor further to get,


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This implies that,


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