Question

If $f\left(x\right)=a{x}^8+b{x}^5+cx+d$, find the value of $f(x+h)-f(x-h)$.

Answer 1

$f\left(x\right)=a{x}^8+b{x}^5+cx+d$

In $f(x+h)-f(x-h)$, using the binomial expansion, terms with even power of $h$ will cancel out and terms with of power of $h$ will add up$f(x-4)=(x-4{)}^4+16(x-4{)}^3+72(x-4{)}^2+64(x-4)-129$

$f(x+h)-f(x-h)=2a\left[{}^8{C}_1{x}^{7}h{+}^{8}C3x^5{h}^3{+}^8{C}_5{x}^3{h}^5{+}^8{C}_7{x}^1{h}^7\right]$

$+2b\left[{}^5{C}_1{x}^{4}h{+}^{5}C3x^2{h}^3{+}^5{C}_5{h}^5\right]+2ch$

$=2[8ah{x}^7+56a{h}^3{x}^5+5bh{x}^4+56a{h}^5{x}^3+10b{h}^3{x}^2+8a{h}^{7}x+10b{h}^5+ch]$