Question

Replace ∗ with a monomial so that the derived equality will be an identity: (15y + ∗)2 = 225y2+12x3y+0.16x6

Answer 1

To make the equation an identity, the asterisk in (15y + *)^2 must be replaced with a monomial that, when squared, produces the given terms on the right side. By matching terms after expanding, the monomial that satisfies the equation is found to be 0.4x^3. Hence, replacing * with 0.4x^3 makes the equation an identity.

Step-by-step explanation:

The student is asking to find a monomial that can be substituted for the asterisk (*) in the given equation so that the resulting expression is an identity. Solving for the asterisk requires understanding how to expand binomials and equating coefficients.

To begin, let's expand the left side of the equation (15y + *)2 as (15y + *) * (15y + *).

Next, using the FOIL method for binomials (First, Outside, Inside, Last), the expanded form would be:

• First: (15y)*(15y) = 225y2
• Outside: (15y)*(*)
• Inside: (*)*(15y)
• Last: (*)*(*)

When we add up these parts, we get:

225y2 + 15y* + 15y* + *2

Now, since the right side of the equation is 225y2 + 12x3y + 0.16x6, we need to match terms.

The term with y2 already matches, so we look at the x3y term. For this term to be a result of the expansion, * must be 0.4x3. This gives us:

• (15y)*(0.4x3) + (0.4x3)*(15y) = (6x3y) + (6x3y) = 12x3y
• (0.4x3)*(0.4x3) = 0.16x6

Hence, the monomial that satisfies the equation is * = 0.4x3.

Answer 2

∗ = 0.4x³

Step-by-step explanation:

(15y + ∗)² = 225y²+12x³y+0.16x⁶

Note:

225y² = 15y * 15y = (15y)²

12x³y = 2 * 15y * 0.4x³

0.16x⁶ = 0.4x³ * 0.4x³ = (0.4x³)²

So, by factoring the right hand side:

225y²+12x³y+0.16x⁶ = (15y + 0.4x³)²

By comparing the left hand side with (15y + 0.4x³)²

So, ∗ should be replaced with the monomial 0.4x³