# Answer to Question #6239 in Algebra for Justin

Question #6239

2

5a + 13a + 8

factor completely

5a + 13a + 8

factor completely

Expert's answer

Let's factor the expression by separating the complete square:

5a^2+13a+8=5(a^2+2a+1)+3a+3=5(a+1)^2+3(a+1)=(a+1)(5(a+1)+3)=(a+1)(5a+5+3)=(a+1)(5a+8)

Also we can do it in another way by finding roots of the given polynome:

5a^2+13a+8=0

D=13^2-4*8*5=169-160=9

a1=(-13+3)/10=-1

a2=(-13-3)/10=-16/10

Now we can factor the polynome as following:

5a^2+13a+8=5(a+1)(a+16/10)=(a+1)(5a+8) - no surprises, we've obtained the same result

For our visitors we also created video where factorisation of quadratic polynom is considered. Please take a look!

5a^2+13a+8=5(a^2+2a+1)+3a+3=5(a+1)^2+3(a+1)=(a+1)(5(a+1)+3)=(a+1)(5a+5+3)=(a+1)(5a+8)

Also we can do it in another way by finding roots of the given polynome:

5a^2+13a+8=0

D=13^2-4*8*5=169-160=9

a1=(-13+3)/10=-1

a2=(-13-3)/10=-16/10

Now we can factor the polynome as following:

5a^2+13a+8=5(a+1)(a+16/10)=(a+1)(5a+8) - no surprises, we've obtained the same result

For our visitors we also created video where factorisation of quadratic polynom is considered. Please take a look!

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