Question

B2+16b+64 -3k3+15k2-6k

Answer 1

Answer:

1. $(b+8)^{2}$

2. $-3k(k^{2} -5k+2)$

3. $(3x-1)(x-5)$

Step-by-step explanation:

1. Both $x^{2}$ and 64 are perfect squares, meaning $16b$ is twice the product of x and 8. Since all signs are positive, the equation would be: $(a+b)^{2} =a^{2} +2ab+b^{2}$. Let $a=x$ and $b=8$. Answer: $(b+8)^{2}$ or $(b+8)(b+8)$.

2. Since -3 is a factor of all 3 terms, factor out the -3 which makes,  $-3(k^{2} +5k^{2} -2k)$ . K is also a common factor, so you would factor that out too, $-3k(k^{2} -5k-2)$. Then simply, find the two factors whose product is -2 and whose sum is -5. Answer: $-3k(k^{2} -5k+2)$.

3. By factoring $3x^{2} -16x+5$, you would break the expressions in the group=$(3x^{2} -x)+(-15x+5)$. Then, factor out "x" from $3x^{2} -x: x(3x-1)$. Factor out "5" from $-15x+5:-5(3x-1)$ = $x(3x-1)-5(3x-1)$. Finally, factor out the common term "3x-1" = $(3x-1)(x-5)$