Question

Factor each completely. If the polynomial is not factorable write prime.

A. b^2 -12b +36

B. 9c^2 -12c +4

C. p^2 -2p-15

D. c^9 +125

Answer 1

\[(x-y)^2= x^2-2xy+y^2\]
this formula works for A and B

Answer 2

how would I use it ?

Answer 3

A.\[ b^2 -12b +36\\=b^2-2\cdot b \cdot 6+6^2\\=\left( b-6 \right)^2\]

Answer 4

B.\[9c^2-12c+4\\=\left( 3c \right)^2-2\left( 3c \right)\cdot2+2^2\\=\left( 3c-2 \right)^2\]

Answer 5

could you show me with C. how you do that?

Answer 6

\[p^2-2p-15\\=p^2-5p+3p-15\]

you rewrite -2p as -5p +3p .... which still adds up to -2p but multiplies to equal the product of the first and last term: \(-15p^2)

Answer 7

then you factor out what's common in the first two terms and the last two terms\[=p^2-5p+3p-15\\=p \left( p-5 \right)+3\left( p-5 \right)\]

Answer 8

and then you factor out the (p-5)\[\large=\left( p-5 \right)\left( p+3 \right)\]

Answer 9

and then your left with both of those?

Answer 10

\[p^2 -2p-15\\=(p-5)(p+3)\]
yup you have the factored form

Answer 11

for D, use the formula:\[x^3+y^3=(x+y)\left( x^2-xy+y^2 \right)\]

Answer 12

I really no good with this formula stuff lol sorry what would i do with it after i write it out

Answer 13

\[\large {c^9+125\\=\left( c^3 \right)^3+5^3}\]

Answer 14

\[\large=\left( c^3+5 \right)\left[\left( c^3 \right)^2-\left( c^3 \right)\cdot5+5^2\right]\]

Answer 15

\[\large=\left( c^3+5 \right)\left( c^6-5c^3+25 \right)\]

and we can't factor further

Answer 16

awsome thanks so much :)