Replace ∗ with a monomial so that the derived equality will be an identity: (15y + ∗)^2 = 225y^2+12x^3y+0.16x^6

Step-by-step

We have :

2×(0,4)×15=12

(0,4)^2=0,16

(15y + ∗)^2 = 225y² + 12x³y + 0.16x⁶

                 = (15y)² + 2×(15y)×(0.4x³) + (0.4x³)²

then

∗ = 0.4x³

Step-by-step explanation:

(a + b)² = a² + 2ab + b²

a = 15y

2ab = 12x³y  

2* 15y*b = 12x³y  

(15y + 0.4x³)

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step-by-step explanation: