( b)(8y^2−4y^4)=40y^4−20y^6 find b
To find the value of b, we need to solve the equation (b)(8y^2−4y^4) = 40y^4−20y^6.
Expanding the left side of the equation:
b(8y^2−4y^4) = 8by^2 − 4by^4
Now we set the left side equal to the right side:
8by^2 − 4by^4 = 40y^4 − 20y^6
Rearranging terms:
4by^4 + 20y^6 = 8by^2
Dividing both sides by 4y^2:
b + 5y^4 = 2b
Subtracting 5y^4 from both sides:
b = 2b - 5y^4
Subtracting b from both sides:
0 = b - 5y^4
Therefore, the value of b is 5y^4.