Question

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Explanation:

46. Step1: Factor the quadratic

$x^2+2x-3=(x+3)(x-1)$

46. Step2: Solve for critical points

$x+3=0 \implies x=-3$; $x-1=0 \implies x=1$

46. Step3: Test intervals for inequality

Test $x<-3$: $(-)(-)=+>0$; Test $-31$: $(+)(+)=+>0$
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47. Step1: Rearrange numerator terms

$6a^2+2a-28$

47. Step2: Factor out GCF from numerator

$2(3a^2+a-14)$

47. Step3: Factor quadratic in numerator

$2(3a+7)(a-2)$

47. Step4: Cancel common factor

$\frac{2(3a+7)(a-2)}{3a+7}=2(a-2)$
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48. Step1: Rearrange numerator terms

$2b^4-b^3+8b+15$

48. Step2: Polynomial long division

Divide $2b^4-b^3+0b^2+8b+15$ by $2b^2-3b+5$:
First term: $\frac{2b^4}{2b^2}=b^2$
Multiply divisor: $b^2(2b^2-3b+5)=2b^4-3b^3+5b^2$
Subtract: $(2b^4-b^3+8b+15)-(2b^4-3b^3+5b^2)=2b^3-5b^2+8b+15$
Next term: $\frac{2b^3}{2b^2}=b$
Multiply divisor: $b(2b^2-3b+5)=2b^3-3b^2+5b$
Subtract: $(2b^3-5b^2+8b+15)-(2b^3-3b^2+5b)=-2b^2+3b+15$
Next term: $\frac{-2b^2}{2b^2}=-1$
Multiply divisor: $-1(2b^2-3b+5)=-2b^2+3b-5$
Subtract: $(-2b^2+3b+15)-(-2b^2+3b-5)=20$
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49. Step1: Rewrite numerator as difference of squares

$1-(9x^2)^2=(1-9x^2)(1+9x^2)$

49. Step2: Factor first term again

$(1-3x)(1+3x)(1+9x^2)$

49. Step3: Cancel common factor

$\frac{(1-3x)(1+3x)(1+9x^2)}{3x+1}=(1-3x)(1+9x^2)$
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50. Step1: Polynomial long division

Divide $8x^3+0x^2-4x+1$ by $2x-1$:
First term: $\frac{8x^3}{2x}=4x^2$
Multiply divisor: $4x^2(2x-1)=8x^3-4x^2$
Subtract: $(8x^3-4x+1)-(8x^3-4x^2)=4x^2-4x+1$
Next term: $\frac{4x^2}{2x}=2x$
Multiply divisor: $2x(2x-1)=4x^2-2x$
Subtract: $(4x^2-4x+1)-(4x^2-2x)=-2x+1$
Next term: $\frac{-2x}{2x}=-1$
Multiply divisor: $-1(2x-1)=-2x+1$
Subtract: $(-2x+1)-(-2x+1)=0$

Answer:

1. $(-\infty, -3) \cup (1, \infty)$
2. $2a-4$
3. $b^2+b-1+\frac{20}{2b^2-3b+5}$
4. $1-3x+9x^2-27x^3$
5. $4x^2+2x-1$