Answer :
Certainly! Let's expand the expression [tex]\((x - 2)^8\)[/tex] step by step.
To expand [tex]\((x - 2)^8\)[/tex], we use the binomial theorem. The binomial theorem states that:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In this case, [tex]\(a = x\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(n = 8\)[/tex]. So, we substitute these values into the binomial theorem:
[tex]\[ (x - 2)^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} (-2)^k \][/tex]
Now, let's compute each term in the sum:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{8}{0} x^{8-0} (-2)^0 = 1 \cdot x^8 \cdot 1 = x^8 \][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{8}{1} x^{8-1} (-2)^1 = 8 \cdot x^7 \cdot (-2) = -16x^7 \][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{8}{2} x^{8-2} (-2)^2 = 28 \cdot x^6 \cdot 4 = 112x^6 \][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{8}{3} x^{8-3} (-2)^3 = 56 \cdot x^5 \cdot (-8) = -448x^5 \][/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]\[ \binom{8}{4} x^{8-4} (-2)^4 = 70 \cdot x^4 \cdot 16 = 1120x^4 \][/tex]
6. For [tex]\(k = 5\)[/tex]:
[tex]\[ \binom{8}{5} x^{8-5} (-2)^5 = 56 \cdot x^3 \cdot (-32) = -1792x^3 \][/tex]
7. For [tex]\(k = 6\)[/tex]:
[tex]\[ \binom{8}{6} x^{8-6} (-2)^6 = 28 \cdot x^2 \cdot 64 = 1792x^2 \][/tex]
8. For [tex]\(k = 7\)[/tex]:
[tex]\[ \binom{8}{7} x^{8-7} (-2)^7 = 8 \cdot x \cdot (-128) = -1024x \][/tex]
9. For [tex]\(k = 8\)[/tex]:
[tex]\[ \binom{8}{8} x^{8-8} (-2)^8 = 1 \cdot 1 \cdot 256 = 256 \][/tex]
Adding all these terms together, we get:
[tex]\[ x^8 - 16x^7 + 112x^6 - 448x^5 + 1120x^4 - 1792x^3 + 1792x^2 - 1024x + 256 \][/tex]
So, the expanded form of [tex]\((x - 2)^8\)[/tex] is:
[tex]\[ x^8 - 16x^7 + 112x^6 - 448x^5 + 1120x^4 - 1792x^3 + 1792x^2 - 1024x + 256 \][/tex]
To expand [tex]\((x - 2)^8\)[/tex], we use the binomial theorem. The binomial theorem states that:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In this case, [tex]\(a = x\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(n = 8\)[/tex]. So, we substitute these values into the binomial theorem:
[tex]\[ (x - 2)^8 = \sum_{k=0}^{8} \binom{8}{k} x^{8-k} (-2)^k \][/tex]
Now, let's compute each term in the sum:
1. For [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{8}{0} x^{8-0} (-2)^0 = 1 \cdot x^8 \cdot 1 = x^8 \][/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{8}{1} x^{8-1} (-2)^1 = 8 \cdot x^7 \cdot (-2) = -16x^7 \][/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{8}{2} x^{8-2} (-2)^2 = 28 \cdot x^6 \cdot 4 = 112x^6 \][/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{8}{3} x^{8-3} (-2)^3 = 56 \cdot x^5 \cdot (-8) = -448x^5 \][/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]\[ \binom{8}{4} x^{8-4} (-2)^4 = 70 \cdot x^4 \cdot 16 = 1120x^4 \][/tex]
6. For [tex]\(k = 5\)[/tex]:
[tex]\[ \binom{8}{5} x^{8-5} (-2)^5 = 56 \cdot x^3 \cdot (-32) = -1792x^3 \][/tex]
7. For [tex]\(k = 6\)[/tex]:
[tex]\[ \binom{8}{6} x^{8-6} (-2)^6 = 28 \cdot x^2 \cdot 64 = 1792x^2 \][/tex]
8. For [tex]\(k = 7\)[/tex]:
[tex]\[ \binom{8}{7} x^{8-7} (-2)^7 = 8 \cdot x \cdot (-128) = -1024x \][/tex]
9. For [tex]\(k = 8\)[/tex]:
[tex]\[ \binom{8}{8} x^{8-8} (-2)^8 = 1 \cdot 1 \cdot 256 = 256 \][/tex]
Adding all these terms together, we get:
[tex]\[ x^8 - 16x^7 + 112x^6 - 448x^5 + 1120x^4 - 1792x^3 + 1792x^2 - 1024x + 256 \][/tex]
So, the expanded form of [tex]\((x - 2)^8\)[/tex] is:
[tex]\[ x^8 - 16x^7 + 112x^6 - 448x^5 + 1120x^4 - 1792x^3 + 1792x^2 - 1024x + 256 \][/tex]
