Question

Factor completely. If a polynomial is prime, state this.

9t^2 + 14t + 5

Answer 1

(1 + t)(5 + 9t)

Answer 2

lets see if i understand "factor by grouping" take
\[9\times 5=45\] then i want two numbers whose product is 45 and whose sum is 14

Answer 3

those would be 9 and 5 because
\[9+5=14\] and \[9\times 5=45\]

Answer 4

9t^(2)+14t+5

Find the factors such that the product of the factors is the trinomial 9t^(2)+14t+5. This can be done by trial and error and checked using the FOIL method of simplifying polynomials.
(t+1)(9t+5)

Answer 5

then write
\[9t^2+9t+5t+5\]
\[9t(t+1)+5(t+1)=(9t+5)(t+1)\]

Answer 6

First get the product of 9 x 5 which gives us 45 Now find two positive factors of 45 whose sum is 14.
9 x 5 is our factors and the sum of 9 + 5 is 14 so now we use the grouping method
9t^2 + 14t + 5
9t^2 + 9t+ 5t + 5

Now Group
(9t^2 + 9t)/9t+ (5t + 5)/5
9t(t+1)+ 5(t+1) = (9t+5)(t+1)

Man I am slow at responding lol

Answer 7

same answer as what I had