Question

trinomial factoring (continued)
factor the following expressions using trinomial \abc\ factoring.

1. ( x^2 + 12x - 13 )
2. ( x^2 - 2x - 8 )
3. ( x^2 + x - 30 )

ready for more?
factor out the gcf of the following.
e1) ( 3x^2 + 9x^{11} )
e2) ( 20g^9 + 30g^6h )
e3) ( x^2y^5z + xy^{22}z )

Explanation:

Step1: Factor $x^2+12x-13$

Find two numbers: $13$ and $-1$, since $13+(-1)=12$ and $13\times(-1)=-13$.
$x^2+12x-13=(x+13)(x-1)$

Step2: Factor $x^2-2x-8$

Find two numbers: $-4$ and $2$, since $-4+2=-2$ and $-4\times2=-8$.
$x^2-2x-8=(x-4)(x+2)$

Step3: Factor $x^2+x-30$

Find two numbers: $6$ and $-5$, since $6+(-5)=1$ and $6\times(-5)=-30$.
$x^2+x-30=(x+6)(x-5)$

Step4: Factor out GCF of $3x^2+9x^{11}$

GCF of $3x^2$ and $9x^{11}$ is $3x^2$.
$3x^2+9x^{11}=3x^2(1+3x^9)$

Step5: Factor out GCF of $20g^9+30g^6h$

GCF of $20g^9$ and $30g^6h$ is $10g^6$.
$20g^9+30g^6h=10g^6(2g^3+3h)$

Step6: Factor out GCF of $x^2y^5z+xy^{22}z$

GCF of $x^2y^5z$ and $xy^{22}z$ is $xy^5z$.
$x^2y^5z+xy^{22}z=xy^5z(x+y^{17})$

Answer:

1. $(x+13)(x-1)$
2. $(x-4)(x+2)$
3. $(x+6)(x-5)$

E1) $3x^2(1+3x^9)$
E2) $10g^6(2g^3+3h)$
E3) $xy^5z(x+y^{17})$