Question

1. try factoring the following by educated guessing: 11x² + 19x - 6
2. create two factoring problems similar to the ones on this worksheet and show their factorizations:
3. now that we have studied some tools for factoring:

a) reflect: what does it mean to factor an expression? what is the goal of factoring?
b) summarize the methods for factoring by grouping and for factoring trinomials (expressions with 3 terms) that we have studied so far. explain/summarize as if you were teaching a peer.

Explanation:

1. For \(11x^{2}+19x - 6\):

• Step1: Find two numbers
• We need two numbers that multiply to \(11\times(-6)=-66\) and add up to \(19\). The numbers are \(22\) and \(-3\) since \(22\times(-3)=-66\) and \(22+( - 3)=19\).
• Step2: Rewrite the middle - term
• \(11x^{2}+19x - 6=11x^{2}+22x-3x - 6\)
• Step3: Group the terms
• \((11x^{2}+22x)-(3x + 6)\)
• Step4: Factor out GCF from each group
• \(11x(x + 2)-3(x + 2)\)
• Step5: Factor out the common binomial factor
• \((11x - 3)(x + 2)\)

1. For the similar factoring problems:

• For \(5x^{2}+13x - 6\):
• Step1: Find two numbers
• We need two numbers that multiply to \(5\times(-6)=-30\) and add up to \(13\). The numbers are \(15\) and \(-2\).
• Step2: Rewrite the middle - term
• \(5x^{2}+13x - 6=5x^{2}+15x-2x - 6\)
• Step3: Group the terms
• \((5x^{2}+15x)-(2x + 6)\)
• Step4: Factor out GCF from each group
• \(5x(x + 3)-2(x + 3)\)
• Step5: Factor out the common binomial factor
• \((5x - 2)(x + 3)\)
• For \(3x^{2}+10x - 8\):
• Step1: Find two numbers
• We need two numbers that multiply to \(3\times(-8)=-24\) and add up to \(10\). The numbers are \(12\) and \(-2\).
• Step2: Rewrite the middle - term
• \(3x^{2}+10x - 8=3x^{2}+12x-2x - 8\)
• Step3: Group the terms
• \((3x^{2}+12x)-(2x + 8)\)
• Step4: Factor out GCF from each group
• \(3x(x + 4)-2(x + 4)\)
• Step5: Factor out the common binomial factor
• \((3x - 2)(x + 4)\)

1. For part a):

• Factoring simplifies expressions and helps in solving equations. For example, if \(ax^{2}+bx + c=(mx + p)(nx+q)\), and we want to solve \(ax^{2}+bx + c = 0\), we can use the zero - product property \((mx + p)(nx + q)=0\) which gives \(mx + p = 0\) or \(nx+q = 0\).
• For part b):
• Factoring by grouping works by splitting the polynomial into groups and finding common factors within and between the groups. For trinomials, we use the product - sum method to rewrite the middle - term and then factor by grouping.

Answer:

1. For \(11x^{2}+19x - 6\):

• The factored form is \((11x - 3)(x+2)\)

1. Two similar factoring - problems and their factorizations:

• Problem 1: \(5x^{2}+13x - 6\)
• Factored form: \((5x - 2)(x + 3)\)
• Explanation: We need to find two numbers that multiply to \(5\times(-6)=-30\) and add up to \(13\). The numbers are \(15\) and \(-2\). Then we rewrite the middle - term \(13x = 15x-2x\) and factor by grouping.
• Problem 2: \(3x^{2}+10x - 8\)
• Factored form: \((3x - 2)(x + 4)\)
• Explanation: We find two numbers that multiply to \(3\times(-8)=-24\) and add up to \(10\). The numbers are \(12\) and \(-2\). Then we rewrite the middle - term \(10x=12x - 2x\) and factor by grouping.

1. a) Factoring an expression means rewriting the given algebraic expression as a product of simpler algebraic expressions. The goal of factoring is to simplify algebraic expressions, solve equations (by setting the factored form equal to zero and using the zero - product property), and understand the structure of the expressions better.

b) Factoring by grouping:

• For a polynomial with four terms, say \(ax^{3}+bx^{2}+cx + d\). First, group the terms into two pairs, like \((ax^{3}+bx^{2})+(cx + d)\). Then factor out the greatest common factor from each pair. For example, if from the first pair we can factor out \(x^{2}\) getting \(x^{2}(ax + b)\) and from the second pair we can factor out \(k\) getting \(k( \frac{c}{k}x+\frac{d}{k})\), and if \((ax + b)\) is common, we can factor it out further.
• Factoring trinomials (\(ax^{2}+bx + c\)):
• First, find two numbers \(m\) and \(n\) such that \(m\times n=ac\) and \(m + n=b\). Then rewrite the middle - term \(bx\) as \(mx+nx\). So the trinomial becomes \(ax^{2}+mx+nx + c\). Group the terms: \((ax^{2}+mx)+(nx + c)\). Factor out the GCF from each group and then factor out the common binomial factor.