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Answer :
To factor the trinomial [tex]\(6x^6 - 60x^5 + 126x^4\)[/tex], let's go through the process step-by-step:
1. Identify the Greatest Common Factor (GCF):
- First, examine the coefficients 6, 60, and 126. The greatest common factor of these numbers is 6.
- In terms of the variable [tex]\(x\)[/tex], the smallest power in all terms is [tex]\(x^4\)[/tex].
- Therefore, the GCF of the entire trinomial is [tex]\(6x^4\)[/tex].
2. Factor Out the GCF:
- Divide each term by the GCF [tex]\(6x^4\)[/tex]:
- [tex]\(6x^6 \div 6x^4 = x^2\)[/tex]
- [tex]\(-60x^5 \div 6x^4 = -10x\)[/tex]
- [tex]\(126x^4 \div 6x^4 = 21\)[/tex]
- Factoring out [tex]\(6x^4\)[/tex] gives us:
[tex]\[
6x^4(x^2 - 10x + 21)
\][/tex]
3. Factor the Remaining Quadratic:
- Now, focus on the quadratic part: [tex]\(x^2 - 10x + 21\)[/tex].
- To factor this, we need two numbers that multiply to 21 and add to -10.
- The pairs of factors for 21 are (1, 21) and (3, 7). None of these pairs add up to -10.
- Therefore, the quadratic [tex]\(x^2 - 10x + 21\)[/tex] does not factor neatly over the integers.
4. Conclusion:
- Since we can’t further factor [tex]\(x^2 - 10x + 21\)[/tex], the fully factored form of the original trinomial is:
[tex]\[
6x^4 (x^2 - 10x + 21)
\][/tex]
- If a quadratic cannot be factored easily or exactly using integers, it is considered "not further factorable" over the integers.
This gives us the factored form of the trinomial: [tex]\(6x^4(x^2 - 10x + 21)\)[/tex]. The trinomial itself is "Not Factorable" over the integers beyond this point.
1. Identify the Greatest Common Factor (GCF):
- First, examine the coefficients 6, 60, and 126. The greatest common factor of these numbers is 6.
- In terms of the variable [tex]\(x\)[/tex], the smallest power in all terms is [tex]\(x^4\)[/tex].
- Therefore, the GCF of the entire trinomial is [tex]\(6x^4\)[/tex].
2. Factor Out the GCF:
- Divide each term by the GCF [tex]\(6x^4\)[/tex]:
- [tex]\(6x^6 \div 6x^4 = x^2\)[/tex]
- [tex]\(-60x^5 \div 6x^4 = -10x\)[/tex]
- [tex]\(126x^4 \div 6x^4 = 21\)[/tex]
- Factoring out [tex]\(6x^4\)[/tex] gives us:
[tex]\[
6x^4(x^2 - 10x + 21)
\][/tex]
3. Factor the Remaining Quadratic:
- Now, focus on the quadratic part: [tex]\(x^2 - 10x + 21\)[/tex].
- To factor this, we need two numbers that multiply to 21 and add to -10.
- The pairs of factors for 21 are (1, 21) and (3, 7). None of these pairs add up to -10.
- Therefore, the quadratic [tex]\(x^2 - 10x + 21\)[/tex] does not factor neatly over the integers.
4. Conclusion:
- Since we can’t further factor [tex]\(x^2 - 10x + 21\)[/tex], the fully factored form of the original trinomial is:
[tex]\[
6x^4 (x^2 - 10x + 21)
\][/tex]
- If a quadratic cannot be factored easily or exactly using integers, it is considered "not further factorable" over the integers.
This gives us the factored form of the trinomial: [tex]\(6x^4(x^2 - 10x + 21)\)[/tex]. The trinomial itself is "Not Factorable" over the integers beyond this point.
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