Question

steps

1. split the 4 terms into two groups
2. factor out the gcf of each pair. note: if the first term of a pair is negative factor out the negative too. if there’s no gcf it’s 1.
3. factor out the common binomial factor note: this is the matching expression in the ()’s

check your understanding
in exercises 1 - 8, factor the polynomial by grouping.

1. $b^{3}-4b^{2}+b - 4$
2. $d^{2}+2c+cd + 2d$
3. $5t^{3}+6t^{2}+5t + 6$
4. $8s^{3}+s-64s^{2}-8$
5. $12a^{3}+2a^{2}-30a - 5$
6. $21h^{3}+18h^{2}-35h - 30$
7. $6x^{2}+9x + 4x+6$
8. $8x^{2}+4x + 2x + 1$

Explanation:

Step1: Group the terms

For \(d^{2}+2c + cd+2d\), group it as \((d^{2}+cd)+(2c + 2d)\).

Step2: Factor out GCF from each pair

From \((d^{2}+cd)\), the GCF is \(d\), so we get \(d(d + c)\). From \((2c + 2d)\), the GCF is \(2\), so we get \(2(c + d)=(d + c)\times2\) (since \(c + d=d + c\)).

Step3: Factor out the common binomial factor

We have \(d(d + c)+2(d + c)=(d + c)(d + 2)\)

Step1: Group the terms

For \(5t^{3}+6t^{2}+5t + 6\), group it as \((5t^{3}+5t)+(6t^{2}+6)\).

Step2: Factor out GCF from each pair

From \((5t^{3}+5t)\), the GCF is \(5t\), so we get \(5t(t^{2}+1)\). From \((6t^{2}+6)\), the GCF is \(6\), so we get \(6(t^{2}+1)\).

Step3: Factor out the common binomial factor

We have \(5t(t^{2}+1)+6(t^{2}+1)=(t^{2}+1)(5t + 6)\)

Step1: Group the terms

For \(8s^{3}+s-64s^{2}-8\), group it as \((8s^{3}-64s^{2})+(s - 8)\).

Step2: Factor out GCF from each pair

From \((8s^{3}-64s^{2})\), the GCF is \(8s^{2}\), so we get \(8s^{2}(s - 8)\). From \((s - 8)\), the GCF is \(1\), so it remains \(s - 8\).

Step3: Factor out the common binomial factor

We have \(8s^{2}(s - 8)+1(s - 8)=(s - 8)(8s^{2}+1)\)

Step1: Group the terms

For \(12a^{3}+2a^{2}-30a - 5\), group it as \((12a^{3}+2a^{2})-(30a + 5)\).

Step2: Factor out GCF from each pair

From \((12a^{3}+2a^{2})\), the GCF is \(2a^{2}\), so we get \(2a^{2}(6a + 1)\). From \((30a + 5)\), the GCF is \(5\), so we get \(5(6a+1)\).

Step3: Factor out the common binomial factor

We have \(2a^{2}(6a + 1)-5(6a + 1)=(6a + 1)(2a^{2}-5)\)

Answer:

\((d + c)(d + 2)\)