Question

expansion of 4x^2+5x+12 in power of (x-1) is​

Answer 1

Answer:

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Answer 2

$4x^{2} +5x+12$ = $21+13(x-1) +4$$(x-1)^{2}$

Given:

$4x^{2} +5x+12$ and we have to expand it in power of (x-1)

we will expand it using the Taylor's theorem, that is

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

Here,

f(x) = $4x^{2} +5x+12$

f(1) = 4 +5+12 = 21

f'(x) = 8x +5

f'(1) = 8 + 5 = 13

f''(x) = 8

f''(1) = 8

Now, putting the values in Taylor's theorem equation

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

$4x^{2} +5x+12$ = $21 + (x-1)13+ \frac{(x-1)}{2!} ^{2} 8$

$4x^{2} +5x+12$ = $21+13(x-1) +4$$(x-1)^{2}$

Hence, the expansion of $4x^{2} +5x+12$ in power of (x-1) is​ $21+13(x-1) +4$$(x-1)^{2}$