Question

chapter 2.4 factoring
gcf out of each polynomial.
x² + 8x

1. 3x⁵y + 27x³y²
2. 7x⁷ + 14x⁴ + 28x

omial completely.

1. x² + 10x + 21
2. x² + 9x + 20

1. x² - 10x + 24
2. x² + 2x - 48

1. 2x² + 5x + 2
2. 4x² + 4x - 3

1. 8x² + 8x - 48
2. 4x² - 19x - 5

Explanation:

Let's solve problem 6: \(x^{2}+9x + 20\)

Step 1: Find two numbers

We need two numbers that multiply to \(20\) (the constant term) and add up to \(9\) (the coefficient of \(x\)). The numbers are \(4\) and \(5\) since \(4\times5 = 20\) and \(4 + 5=9\).

Step 2: Factor the quadratic

Using the numbers found, we can factor the quadratic as \((x + 4)(x+5)\) because when we expand \((x + 4)(x + 5)\), we get \(x^{2}+5x+4x + 20=x^{2}+9x + 20\).

Now problem 5: \(x^{2}+10x + 21\)

Step 1: Find two numbers

We need two numbers that multiply to \(21\) and add up to \(10\). The numbers are \(3\) and \(7\) since \(3\times7 = 21\) and \(3+7 = 10\).

Step 2: Factor the quadratic

Factor it as \((x + 3)(x + 7)\) (expanding \((x + 3)(x+7)=x^{2}+7x+3x + 21=x^{2}+10x + 21\)).

Problem 8: \(x^{2}-10x + 24\)

Step 1: Find two numbers

We need two numbers that multiply to \(24\) and add up to \(- 10\) (since the coefficient of \(x\) is \(-10\) and the constant term is positive, both numbers are negative). The numbers are \(-4\) and \(-6\) since \((-4)\times(-6)=24\) and \(-4+(-6)=-10\).

Step 2: Factor the quadratic

Factor it as \((x - 4)(x - 6)\) (expanding \((x - 4)(x - 6)=x^{2}-6x-4x + 24=x^{2}-10x + 24\)).

Problem 9: \(x^{2}+2x-48\)

Step 1: Find two numbers

We need two numbers that multiply to \(-48\) and add up to \(2\). The numbers are \(8\) and \(-6\) since \(8\times(-6)=-48\) and \(8+(-6)=2\).

Step 2: Factor the quadratic

Factor it as \((x + 8)(x - 6)\) (expanding \((x + 8)(x - 6)=x^{2}-6x+8x-48=x^{2}+2x - 48\)).

Problem 11: \(2x^{2}+5x + 2\)

Step 1: Multiply \(a\) and \(c\)

For a quadratic \(ax^{2}+bx + c\), here \(a = 2\), \(b = 5\), \(c = 2\). Multiply \(a\) and \(c\): \(2\times2=4\).

Step 2: Find two numbers for \(ac\)

Find two numbers that multiply to \(4\) and add up to \(5\). The numbers are \(1\) and \(4\).

Step 3: Rewrite the middle term

Rewrite the middle term \(5x\) as \(x+4x\). So the quadratic becomes \(2x^{2}+x + 4x+2\).

Step 4: Group and factor

Group the first two and last two terms: \((2x^{2}+x)+(4x + 2)=x(2x + 1)+2(2x + 1)\). Then factor out \((2x + 1)\) to get \((2x + 1)(x + 2)\).

Problem 12: \(4x^{2}+4x-3\)

Step 1: Multiply \(a\) and \(c\)

Here \(a = 4\), \(b = 4\), \(c=-3\). Multiply \(a\) and \(c\): \(4\times(-3)=-12\).

Step 2: Find two numbers for \(ac\)

Find two numbers that multiply to \(-12\) and add up to \(4\). The numbers are \(6\) and \(-2\) since \(6\times(-2)=-12\) and \(6+(-2)=4\).

Step 3: Rewrite the middle term

Rewrite \(4x\) as \(6x-2x\). The quadratic becomes \(4x^{2}+6x-2x - 3\).

Step 4: Group and factor

Group: \((4x^{2}+6x)+(-2x - 3)=2x(2x + 3)-1(2x + 3)\). Factor out \((2x + 3)\) to get \((2x + 3)(2x - 1)\).

Problem 14: \(8x^{2}+8x-48\)

Step 1: Factor out GCF

First, factor out the greatest common factor, which is \(8\). So we get \(8(x^{2}+x - 6)\).

Step 2: Factor the quadratic inside

For \(x^{2}+x - 6\), find two numbers that multiply to \(-6\) and add up to \(1\). The numbers are \(3\) and \(-2\). Factor it as \((x + 3)(x - 2)\). So the fully factored form is \(8(x + 3)(x - 2)\).

Problem 15: \(4x^{2}-19x-5\)

Step 1: Multiply \(a\) and \(c\)

Here \(a = 4\), \(b=-19\), \(c = - 5\). Multiply \(a\) and \(c\): \(4\times(-5)=-20\).

Step 2: Find two numbers for \(ac\)

Find two numbers that multiply to \(-20\) and add up to \(-19\). The numbers are \(-20\) and \(1\).

Step 3: Rewrite the middle term

Rewrite \(-19x\) as \(-20x+x\). The quadratic becomes \(4x^{2}-20x+x - 5\).

Step 4: Group and factor

Group: \((4x^{2}-20x)+(x - 5)=4x(x - 5)+1(x - 5)\). Factor out \((x - 5)\) to get \((4x + 1)(x - 5)\).

Problem 2: \(3x^{5}y+27x^{3}y^{2}\)

Step 1: Find GCF of coefficients and variables

The GCF of \(3\) and \(27\) is \(3\). For \(x\) terms, the GCF of \(x^{5}\) and \(x^{3}\) is \(x^{3}\). For \(y\) terms, the GCF of \(y\) and \(y^{2}\) is \(y\). So the GCF is \(3x^{3}y\).

Step 2: Factor out GCF

Factor out \(3x^{3}y\) from the polynomial: \(3x^{3}y(x^{2}+9y)\).

Problem 3: \(7x^{7}+14x^{4}+28x\)

Step 1: Find GCF of coefficients and variables

The GCF of \(7\), \(14\), and \(28\) is \(7\). For \(x\) terms, the GCF of \(x^{7}\), \(x^{4}\), and \(x\) is \(x\). So the GCF is \(7x\).

Step 2: Factor out GCF

Factor out \(7x\) from the polynomial: \(7x(x^{6}+2x^{3}+4)\).

(Note: For the first problem in the image (the one with \(x^{2}+2x + 4\) written, since it's not clear if it's a problem or a note, we skipped it. If it's a problem like factoring \(x^{2}+2x + 4\), it doesn't factor over the real numbers with integer coefficients as the discriminant \(b^{2}-4ac=4 - 16=-12<0\))

Answer:

• Problem 2: \(3x^{3}y(x^{2}+9y)\)
• Problem 3: \(7x(x^{6}+2x^{3}+4)\)
• Problem 5: \((x + 3)(x + 7)\)
• Problem 6: \((x + 4)(x + 5)\)
• Problem 8: \((x - 4)(x - 6)\)
• Problem 9: \((x + 8)(x - 6)\)
• Problem 11: \((2x + 1)(x + 2)\)
• Problem 12: \((2x + 3)(2x - 1)\)
• Problem 14: \(8(x + 3)(x - 2)\)
• Problem 15: \((4x + 1)(x - 5)\)