Just not good at this, factor complete, or state that the polynomial is prime.
9y^4+42y^3+49y^2
(9+y)(9-y)+42(y+3)(y-3)^2+49(y+2(y-2)
I can not seem to get it right. so help me please
9y^4+42y^3+49y^2
always look for a common factor first, sure enough - y^2
= y^2(9y^2 + 42y + 49)
mmmhh? squares in front and back, is it a perfect square? yup!!
= y^2(3y+7)(3y+7)
To determine if the polynomial 9y^4 + 42y^3 + 49y^2 can be factored completely or if it is prime, we need to look for common factors and attempt to factor it further.
First, let's check if there are any common factors among the terms. In this case, we can see that all the terms are divisible by 7, so we can factor out 7:
7(9y^4 + 6y^3 + 7y^2)
Next, let's look for any common factors among the remaining terms. We can see that all the terms have a factor of y^2, so we can factor that out as well:
7y^2(9y^2 + 6y + 7)
Now, we have a quadratic expression left, 9y^2 + 6y + 7. To determine if this quadratic can be factored, we can check if it has any real roots by using the quadratic formula:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the roots (solutions) are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic 9y^2 + 6y + 7, a = 9, b = 6, and c = 7. Plugging these values into the quadratic formula:
y = [-(6) ± √((6)^2 - 4(9)(7))] / (2(9))
y = [-6 ± √(36 - 252)] / 18
y = [-6 ± √(-216)] / 18
Since the term inside the square root (√) is negative, we can see that the quadratic 9y^2 + 6y + 7 does not have real roots. This means that it cannot be factored further using real numbers.
Therefore, the final factored form of the polynomial 9y^4 + 42y^3 + 49y^2 is:
7y^2(9y^2 + 6y + 7)
And we can say that the polynomial is not factorable further using real numbers, making it prime in its factored form.