Answer:
1. [tex](b+8)^{2}[/tex]
2. [tex]-3k(k^{2} -5k+2)[/tex]
3. [tex](3x-1)(x-5)[/tex]
Step-by-step explanation:
1. Both [tex]x^{2}[/tex] and 64 are perfect squares, meaning [tex]16b[/tex] is twice the product of x and 8. Since all signs are positive, the equation would be: [tex](a+b)^{2} =a^{2} +2ab+b^{2}[/tex]. Let [tex]a=x[/tex] and [tex]b=8[/tex]. Answer: [tex](b+8)^{2}[/tex] or [tex](b+8)(b+8)[/tex].
2. Since -3 is a factor of all 3 terms, factor out the -3 which makes, [tex]-3(k^{2} +5k^{2} -2k)[/tex] . K is also a common factor, so you would factor that out too, [tex]-3k(k^{2} -5k-2)[/tex]. Then simply, find the two factors whose product is -2 and whose sum is -5. Answer: [tex]-3k(k^{2} -5k+2)[/tex].
3. By factoring [tex]3x^{2} -16x+5[/tex], you would break the expressions in the group=[tex](3x^{2} -x)+(-15x+5)[/tex]. Then, factor out "x" from [tex]3x^{2} -x: x(3x-1)[/tex]. Factor out "5" from [tex]-15x+5:-5(3x-1)[/tex] = [tex]x(3x-1)-5(3x-1)[/tex]. Finally, factor out the common term "3x-1" = [tex](3x-1)(x-5)[/tex]
