Get expert advice and community support for all your questions on IDNLearn.com. Find the answers you need quickly and accurately with help from our knowledgeable and experienced experts.
Sagot :
Alright, let's go through the step-by-step process of factoring the polynomial [tex]\(3x^3 - 15x^2 + 8x - 40\)[/tex].
1. Group the terms in pairs:
[tex]\[ 3x^3 - 15x^2 + 8x - 40 \Rightarrow (3x^3 - 15x^2) + (8x - 40) \][/tex]
2. Factor out the greatest common factor from each pair:
- For the first group, [tex]\(3x^3 - 15x^2\)[/tex]:
[tex]\[ 3x^2(x - 5) \][/tex]
- For the second group, [tex]\(8x - 40\)[/tex]:
[tex]\[ 8(x - 5) \][/tex]
So we have:
[tex]\[ 3x^2(x - 5) + 8(x - 5) \][/tex]
3. Factor out the common binomial factor [tex]\((x - 5)\)[/tex]:
Notice that [tex]\((x - 5)\)[/tex] is a common factor in both terms. Hence, we can factor it out:
[tex]\[ (x - 5)(3x^2 + 8) \][/tex]
Thus, the completely factored form of the polynomial [tex]\(3x^3 - 15x^2 + 8x - 40\)[/tex] is:
[tex]\[ (x - 5)(3x^2 + 8) \][/tex]
So, the correct answer is:
[tex]\[ \boxed{(3x^2 + 8)(x - 5)} \][/tex]
1. Group the terms in pairs:
[tex]\[ 3x^3 - 15x^2 + 8x - 40 \Rightarrow (3x^3 - 15x^2) + (8x - 40) \][/tex]
2. Factor out the greatest common factor from each pair:
- For the first group, [tex]\(3x^3 - 15x^2\)[/tex]:
[tex]\[ 3x^2(x - 5) \][/tex]
- For the second group, [tex]\(8x - 40\)[/tex]:
[tex]\[ 8(x - 5) \][/tex]
So we have:
[tex]\[ 3x^2(x - 5) + 8(x - 5) \][/tex]
3. Factor out the common binomial factor [tex]\((x - 5)\)[/tex]:
Notice that [tex]\((x - 5)\)[/tex] is a common factor in both terms. Hence, we can factor it out:
[tex]\[ (x - 5)(3x^2 + 8) \][/tex]
Thus, the completely factored form of the polynomial [tex]\(3x^3 - 15x^2 + 8x - 40\)[/tex] is:
[tex]\[ (x - 5)(3x^2 + 8) \][/tex]
So, the correct answer is:
[tex]\[ \boxed{(3x^2 + 8)(x - 5)} \][/tex]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.