Question

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

Answer 1

1.
`(b+8)^(2)`

2.
`-3k(k^(2) -5k+2)`

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

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)`