Question

1. $(y + 3)(2y^2 - 3y + 4)$
2. $(3x + 7)(\frac{2}{7}x^2 - \frac{10}{3}x + 2)$
3. $(2x^2 - 3x)(-3x^2 + 4x - 2)$
4. $(-0.2x^2 + 1)(4x^2 - 2.6x - 3.4)$
5. $(x^2 + 3x)(3x^2 - 2x + 4)$

Explanation:

Let's solve problem 27: \((y + 3)(2y^2 - 3y + 4)\)

Step 1: Distribute \(y\)

Multiply \(y\) with each term in \(2y^2 - 3y + 4\):
\(y\times2y^2 = 2y^3\)
\(y\times(-3y) = -3y^2\)
\(y\times4 = 4y\)
So we get \(2y^3 - 3y^2 + 4y\)

Step 2: Distribute \(3\)

Multiply \(3\) with each term in \(2y^2 - 3y + 4\):
\(3\times2y^2 = 6y^2\)
\(3\times(-3y) = -9y\)
\(3\times4 = 12\)
So we get \(6y^2 - 9y + 12\)

Step 3: Combine like terms

Add the results from Step 1 and Step 2:
\((2y^3 - 3y^2 + 4y) + (6y^2 - 9y + 12)\)
Combine the \(y^2\) terms: \(-3y^2 + 6y^2 = 3y^2\)
Combine the \(y\) terms: \(4y - 9y = -5y\)
The \(y^3\) term and the constant term remain as is.
So we have \(2y^3 + 3y^2 - 5y + 12\)

Step 1: Distribute \(2x^2\)

Multiply \(2x^2\) with each term in \(-3x^2 + 4x - 2\):
\(2x^2\times(-3x^2) = -6x^4\)
\(2x^2\times4x = 8x^3\)
\(2x^2\times(-2) = -4x^2\)
So we get \(-6x^4 + 8x^3 - 4x^2\)

Step 2: Distribute \(-3x\)

Multiply \(-3x\) with each term in \(-3x^2 + 4x - 2\):
\(-3x\times(-3x^2) = 9x^3\)
\(-3x\times4x = -12x^2\)
\(-3x\times(-2) = 6x\)
So we get \(9x^3 - 12x^2 + 6x\)

Step 3: Combine like terms

Add the results from Step 1 and Step 2:
\((-6x^4 + 8x^3 - 4x^2) + (9x^3 - 12x^2 + 6x)\)
Combine the \(x^3\) terms: \(8x^3 + 9x^3 = 17x^3\)
Combine the \(x^2\) terms: \(-4x^2 - 12x^2 = -16x^2\)
The \(x^4\) term and the \(x\) term remain as is.
So we have \(-6x^4 + 17x^3 - 16x^2 + 6x\)

Step 1: Distribute \(x^2\)

Multiply \(x^2\) with each term in \(3x^2 - 2x + 4\):
\(x^2\times3x^2 = 3x^4\)
\(x^2\times(-2x) = -2x^3\)
\(x^2\times4 = 4x^2\)
So we get \(3x^4 - 2x^3 + 4x^2\)

Step 2: Distribute \(3x\)

Multiply \(3x\) with each term in \(3x^2 - 2x + 4\):
\(3x\times3x^2 = 9x^3\)
\(3x\times(-2x) = -6x^2\)
\(3x\times4 = 12x\)
So we get \(9x^3 - 6x^2 + 12x\)

Step 3: Combine like terms

Add the results from Step 1 and Step 2:
\((3x^4 - 2x^3 + 4x^2) + (9x^3 - 6x^2 + 12x)\)
Combine the \(x^3\) terms: \(-2x^3 + 9x^3 = 7x^3\)
Combine the \(x^2\) terms: \(4x^2 - 6x^2 = -2x^2\)
The \(x^4\) term and the \(x\) term remain as is.
So we have \(3x^4 + 7x^3 - 2x^2 + 12x\)

Answer:

\(2y^3 + 3y^2 - 5y + 12\)

Let's solve problem 29: \((2x^2 - 3x)(-3x^2 + 4x - 2)\)