Grade 9 algebra] is their a faster way to do these problems, it ... - Free Printable

To solve the problem of factoring the given polynomials, we will go through each polynomial step by step. Here's a detailed explanation for a few selected problems to illustrate the process:

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Section A: Factoring Quadratic Polynomials

#### Problem 1: \( 3x^2 + 7x + 2 \)

We need to factor \( 3x^2 + 7x + 2 \).

1. Identify coefficients: \( a = 3 \), \( b = 7 \), \( c = 2 \).
2. Find two numbers whose product is \( ac = 3 \cdot 2 = 6 \) and whose sum is \( b = 7 \).
- The numbers are \( 6 \) and \( 1 \) because \( 6 \cdot 1 = 6 \) and \( 6 + 1 = 7 \).
3. Rewrite the middle term: Split \( 7x \) into \( 6x + x \).
\[
3x^2 + 7x + 2 = 3x^2 + 6x + x + 2
\]
4. Factor by grouping:
\[
(3x^2 + 6x) + (x + 2) = 3x(x + 2) + 1(x + 2)
\]
5. Factor out the common binomial:
\[
(3x + 1)(x + 2)
\]

Verification: Multiply the factors to check:
\[
(3x + 1)(x + 2) = 3x^2 + 6x + x + 2 = 3x^2 + 7x + 2
\]
The factorization is correct.

Answer: \( (3x + 1)(x + 2) \)

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#### Problem 3: \( 3c^2 - 8c + 5 \)

We need to factor \( 3c^2 - 8c + 5 \).

1. Identify coefficients: \( a = 3 \), \( b = -8 \), \( c = 5 \).
2. Find two numbers whose product is \( ac = 3 \cdot 5 = 15 \) and whose sum is \( b = -8 \).
- The numbers are \( -3 \) and \( -5 \) because \( (-3) \cdot (-5) = 15 \) and \( -3 + (-5) = -8 \).
3. Rewrite the middle term: Split \( -8c \) into \( -3c - 5c \).
\[
3c^2 - 8c + 5 = 3c^2 - 3c - 5c + 5
\]
4. Factor by grouping:
\[
(3c^2 - 3c) + (-5c + 5) = 3c(c - 1) - 5(c - 1)
\]
5. Factor out the common binomial:
\[
(3c - 5)(c - 1)
\]

Verification: Multiply the factors to check:
\[
(3c - 5)(c - 1) = 3c^2 - 3c - 5c + 5 = 3c^2 - 8c + 5
\]
The factorization is correct.

Answer: \( (3c - 5)(c - 1) \)

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#### Problem 5: \( 5y^2 + 4y - 1 \)

We need to factor \( 5y^2 + 4y - 1 \).

1. Identify coefficients: \( a = 5 \), \( b = 4 \), \( c = -1 \).
2. Find two numbers whose product is \( ac = 5 \cdot (-1) = -5 \) and whose sum is \( b = 4 \).
- The numbers are \( 5 \) and \( -1 \) because \( 5 \cdot (-1) = -5 \) and \( 5 + (-1) = 4 \).
3. Rewrite the middle term: Split \( 4y \) into \( 5y - y \).
\[
5y^2 + 4y - 1 = 5y^2 + 5y - y - 1
\]
4. Factor by grouping:
\[
(5y^2 + 5y) + (-y - 1) = 5y(y + 1) - 1(y + 1)
\]
5. Factor out the common binomial:
\[
(5y - 1)(y + 1)
\]

Verification: Multiply the factors to check:
\[
(5y - 1)(y + 1) = 5y^2 + 5y - y - 1 = 5y^2 + 4y - 1
\]
The factorization is correct.

Answer: \( (5y - 1)(y + 1) \)

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General Approach for Other Problems

For each polynomial, follow these steps:
1. Identify the coefficients \( a \), \( b \), and \( c \).
2. Find two numbers whose product is \( ac \) and whose sum is \( b \).
3. Rewrite the middle term using these two numbers.
4. Factor by grouping.
5. Verify the factorization by multiplying the factors.

If a polynomial cannot be factored, write "prime."

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Final Answer

Here are the factorizations for the first three problems in Section A:

1. \( 3x^2 + 7x + 2 = \boxed{(3x + 1)(x + 2)} \)
2. \( 3c^2 - 8c + 5 = \boxed{(3c - 5)(c - 1)} \)
3. \( 5y^2 + 4y - 1 = \boxed{(5y - 1)(y + 1)} \)

For the remaining problems, apply the same method. If you need help with specific ones, let me know!