Question

the constant term in the expression 49+69x+42x^2+11x^3+x^4 in power of (x+2) is​

Answer 1

Answer:

Answer

the constant term is 2

hope it helps...

Answer 2

The constant term in power of (x+2) is 7

Given:

$49+69x+42x^{2}+11x^{3}+x^{4}$

we will expand the equation $x^{4} +11x^{3}+42x^{2} +69x+49$ in power of (x+2) by using Taylor's theorem

$f(x)=f(a)+(x+a)f'(a)+\frac{(x+a)^{2} }{2!} f''(a)+\frac{(x+a)^{3} }{3!} f'''(a)+\frac{(x+a)^{4} }{4!} f''''(a)$

Here,

f(x)= $x^{4} +11x^{3}+42x^{2} +69x+49$

f(-2) = $(-2)^{4} +11(-2)^{3}+42(-2)^{2} +69(-2)+49$

f(-2) = 16 - 88 + 168 - 138 +49 = 7

f'(x) = $4x^{3} +33x^{2} +84x+69$

f'(-2) = $4(-2)^{3} +33(-2)^{2} +84(-2)+69$

f'(-2) = -32 +132 - 168 + 69 = 1

f''(x) = $12x^{2} +66x+84$

f''(-2) = $12(-2)^{2} +66(-2)+84$

f''(-2) = 48 - 132 + 84 = 0

f'''(x) = $24x+66$

f'''(-2) = 24(-2) + 66 = 18

f''''(x) = 24

f''''(-2) = 24

f'''''(-2)= 0

Now, putting the values in Taylor's theorem equation

$f(x)=f(-2)+(x+2)f'(-2)+\frac{(x+2)^{2} }{2!} f''(-2)+\frac{(x+2)^{3} }{3!} f'''(-2)+\frac{(x+2)^{4} }{4!} f''''(-2)$

f(x) = 7 + (x+2)+$\frac{(x+2)^{2} }{2!}$ 0 + $\frac{(x+2)^{3} }{3!}$18 + $\frac{(x+2)^{4} }{4!}$24

f(x) = 7 + (x+2) +3$(x+2)^{3}$+ $(x+2)^{4}$

Hence, the constant term in power of (x+2) is 7.