Question

expand x^3-7x^2+x-6 in the power of (x-3)​

Answer 1

Step-by-step explanation:

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Answer 2

We have to expand x³ - 7x² + x - 6 in the power of (x - 3).

We can expand, x³ - 7x² + x - 6 in the power of (x - 3) by using Taylor's theorem.

according to Taylor's series,

$f(x)=f(a)+(x+a)f'(a)+\frac{(x+a)^2}{2!}f"(a)+\frac{(x+a)^3}{3!}f'''(a)+...$

let f(x) = x³ - 7x² + x - 6

f(3) = 3³ - 7(3)² + 3 - 6

= 27 - 63 - 3

= -39

first derivative ⇒f'(x) = 3x² - 14x + 1

f'(3) = 3(3)² - 14 × 3 + 1 = 27 - 42 + 1 = -14

2nd derivative ⇒f"(x) = 6x - 14

f"(3) = 6 × 3 - 14 = 4

3rd derivative ⇒f'''(x) = 6

f'''(3) = 6

4th derivative ⇒f''''(x) = 0

f''''(3) = 0

∴ f(x) = f(3) + (x - 3)f'(3) + {(x - 3)²/2!}f"(3) {(x - 3)³/3!}f'''(3) + {(x - 3)⁴/4!}f""(3) + ....

= -39 + (x - 3)(-14) + {(x -3)²/2} × 4 + {(x - 3)³/6} × 6 + {(x - 3)⁴/24} × 0

= -39 - 14(x - 3) + 2(x - 3)² + (x - 3)³

= (x - 3)³ + 2(x - 3)² - 14(x - 3) - 39

Therefore the expansion of x³ - 7x² + x - 6 in the power of (x - 3) is (x - 3)³ + 2(x - 3)² - 14(x - 3) - 39