solve each equation for factoring.\n1. $x^{2}-9x + 20 = 0$\n2. $16a^{2}-1 = 0$\n3. $p^{2}+3p = 0$\n4…

Explanation:

Step1: Factor the first - equation

For the quadratic equation (x^{2}-9x + 20=0), we need to find two numbers that multiply to (20) and add up to (-9). The numbers are (-4) and (-5). So, ((x - 4)(x - 5)=0). Then, by the zero - product property, (x-4 = 0) or (x - 5=0), and (x=4) or (x = 5).

Step2: Factor the second - equation

For the equation (16a^{2}-1=0), we use the difference of squares formula (A^{2}-B^{2}=(A + B)(A - B)), where (A = 4a) and (B = 1). So, ((4a+1)(4a - 1)=0). Then, (4a+1 = 0) gives (a=-\frac{1}{4}), and (4a - 1=0) gives (a=\frac{1}{4}).

Step3: Factor the third - equation

For the equation (p^{2}+3p=0), we factor out the common factor (p). So, (p(p + 3)=0). Then, (p=0) or (p+3 = 0), and (p = 0) or (p=-3).

Step4: Factor the fourth - equation

For the quadratic equation (7x^{2}-13x-2=0), we need to find two numbers that multiply to (7\times(-2)=-14) and add up to (-13). The numbers are (-14) and (1). We rewrite the middle term: (7x^{2}-14x+x - 2=0), then factor by grouping: (7x(x - 2)+1(x - 2)=0), ((7x + 1)(x - 2)=0). So, (7x+1 = 0) gives (x=-\frac{1}{7}), and (x - 2=0) gives (x = 2).

Step5: Rearrange and factor the fifth - equation

First, rewrite (7x^{2}+5x-24=6x^{2}) as (7x^{2}-6x^{2}+5x-24=0), which simplifies to (x^{2}+5x-24=0). We find two numbers that multiply to (-24) and add up to (5). The numbers are (8) and (-3). So, ((x + 8)(x - 3)=0). Then, (x+8 = 0) gives (x=-8), and (x - 3=0) gives (x = 3).

Step6: Rearrange and factor the sixth - equation

Rewrite (x^{2}+1 = 2x) as (x^{2}-2x + 1=0). Using the perfect - square formula ((a - b)^{2}=a^{2}-2ab + b^{2}) with (a=x) and (b = 1), we have ((x - 1)^{2}=0). Then, (x=1).

Answer:

1. (x = 4) or (x = 5)
2. (a=-\frac{1}{4}) or (a=\frac{1}{4})
3. (p = 0) or (p=-3)
4. (x=-\frac{1}{7}) or (x = 2)
5. (x=-8) or (x = 3)
6. (x = 1)