Question

appendix d part 2 worksheet

1. factor each expression, if possible. then, check with distribution.

a) $7x^{2}-7x - 42$
b) $11x + 4x^{2}+6$
c) $3x^{2}+5x - 2$
d) $6w^{2}+4w + 9$

1. similar to some wiley problems decide which rearrangement(s) would help in factoring $2x^{2}-10x - 8$ select all that would work.

a) $3x^{2}-12x + 2x - 8$
b) $2x^{2}+6x - 4x - 8$
c) $2x^{2}+2x - 12x - 8$
d) none of the above since no two integers multiply to - 8 and add to - 10

Explanation:

Step1: Factor 7x² - 7x - 42

First, factor out the greatest - common factor 7: $7(x^{2}-x - 6)$. Then, factor the quadratic $x^{2}-x - 6$. We need two numbers that multiply to - 6 and add up to - 1. These numbers are - 3 and 2. So, $7(x^{2}-x - 6)=7(x - 3)(x+2)$.
Check: Using the distributive property, $7(x - 3)(x + 2)=7(x^{2}+2x-3x - 6)=7(x^{2}-x - 6)=7x^{2}-7x - 42$.

Step2: Factor 4x²+11x + 6

For the quadratic $ax^{2}+bx + c$ (here $a = 4$, $b = 11$, $c = 6$), we need to find two numbers that multiply to $ac=4\times6 = 24$ and add up to $b = 11$. The numbers are 8 and 3. Rewrite the middle term: $4x^{2}+8x+3x + 6$. Group the terms: $(4x^{2}+8x)+(3x + 6)=4x(x + 2)+3(x + 2)=(4x + 3)(x + 2)$.
Check: $(4x + 3)(x + 2)=4x^{2}+8x+3x + 6=4x^{2}+11x + 6$.

Step3: Factor 3x²+5x - 2

For the quadratic $3x^{2}+5x - 2$ where $a = 3$, $b = 5$, $c=-2$, we need two numbers that multiply to $ac=3\times(-2)=-6$ and add up to $b = 5$. The numbers are 6 and - 1. Rewrite the middle term: $3x^{2}+6x - x - 2$. Group the terms: $(3x^{2}+6x)-(x + 2)=3x(x + 2)-(x + 2)=(3x - 1)(x + 2)$.
Check: $(3x - 1)(x + 2)=3x^{2}+6x - x - 2=3x^{2}+5x - 2$.

Step4: Analyze 6w²+4w + 9

For the quadratic $6w^{2}+4w + 9$, where $a = 6$, $b = 4$, $c = 9$, the discriminant $\Delta=b^{2}-4ac=4^{2}-4\times6\times9=16 - 216=-200<0$. So, it cannot be factored over the real - numbers.

Step5: Analyze 2x²-10x - 8

First, factor out the greatest - common factor 2: $2(x^{2}-5x - 4)$. For the quadratic $x^{2}-5x - 4$, the discriminant $\Delta=(-5)^{2}-4\times1\times(-4)=25 + 16 = 41$. Since 41 is not a perfect square, it cannot be factored over the integers.
For the rearrangements:
$3x^{2}-12x+2x - 8=3x^{2}-10x - 8$ (not the same as $2x^{2}-10x - 8$).
$2x^{2}+6x - 4x - 8=2x^{2}+2x - 8$ (not the same as $2x^{2}-10x - 8$).
$2x^{2}+2x-12x - 8=2x^{2}-10x - 8$. So, the rearrangement $2x^{2}+2x-12x - 8$ would help in factoring.

Answer:

a) $7(x - 3)(x + 2)$
b) $(4x + 3)(x + 2)$
c) $(3x - 1)(x + 2)$
d) Cannot be factored over the real - numbers

1. c) $2x^{2}+2x-12x - 8$