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Question Number 128176 by liberty last updated on 05/Jan/21

$$\mathrm{Given}\mathrm{t}\left(\mathrm{x}\right)={\mathrm{ax}}^4+{\mathrm{bx}}^2+\mathrm{x}+5;\mathrm{where}\mathrm{a}\mathrm{and}\mathrm{b}$$$$\mathrm{are}\mathrm{constant}.\mathrm{If}\mathrm{t}(-4)=3\mathrm{then}\mathrm{t}\left(4\right)=?$$

Answered by bemath last updated on 05/Jan/21

$$(\Rightarrow )\mathrm{t}\left(\mathrm{x}\right)={\mathrm{ax}}^4+{\mathrm{bx}}^2+\mathrm{x}+5$$$$(\Rightarrow )\mathrm{t}\left(4\right)=256\mathrm{a}+16\mathrm{b}+9$$$$\mathrm{we}\mathrm{have}\mathrm{t}(-4)=256\mathrm{a}+16\mathrm{b}+1$$$$\mathrm{so}\mathrm{t}(-4)\mathrm{gives}\mathrm{us}256\mathrm{a}+16\mathrm{b}=2$$$$\mathrm{so}\mathrm{we}\mathrm{find}\mathrm{t}\left(4\right)=256\mathrm{a}+16\mathrm{b}+9=2+9=11$$$$$$

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