15v2+52v+45 = (5v+9)(3v+5) when factord
If a multiple of 10 were a factor of 45, then both '10' and the multiple would be factors of 45. Since 10 is not a factor of 45, there is no such number.There are no multiples of 10 that are factors of 45.
360 !
45 +45 ------ 90
There is no multiple of 10 that's a factor of 45.
(t - 15)(t + 3)
The highest common factor of 15 and 45 is 15
One minor correction on your question here - it's not a binomial. A binomial is a polynomial with two terms. This one - having three terms, is a trinomial. To factor it, you need to find two numbers that multiply to make -45 (the last term), and add to make 4 (the coefficient of the middle term). Once you have those numbers, you can break the equation down. In this case the numbers we need are 9 and -5. Here's how it's done: x2 + 4x - 45 = x2 - 5x + 9x - 45 = x(x - 5) + 9(x - 5) = (x + 9)(x - 5)
15v2+52v+45 = (5v+9)(3v+5) when factord
x2 + 18x + 45 = (x + 15)(x + 3).
(x+3)(x+15)
5(4x + 9)
(z + 15)(z + 3)
x2 - 14x + 45 = (x - 5)(x - 9).
5(x^2 + 9)
x2 + 14x + 45 = (x + 9) (x + 5)
(x + 15)(x - 3)