# Find the value of $f(x + 3),$if $f(x) = {x^4} - 12{x^3} + 17{x^2} - 9x + 7.$

Last updated date: 27th Mar 2023

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Answer

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Hint: For finding the value of $f(x + 3),$put $x + 3$in place of $x$in $f(x).$

The given function in the question is $f(x) = {x^4} - 12{x^3} + 17{x^2} - 9x + 7$

For $f(x + 3),$we will replace $x$with $x + 3,$we’ll get:

$ \Rightarrow f(x + 3) = {(x + 3)^4} - 12{(x + 3)^3} + 17{(x + 3)^2} - 9(x + 3) + 7$

For ${(x + 3)^4},$we’ll use binomial expansion and for rest of the terms we use formulae as:

$

{(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) \\

{(a + b)^2} = {a^2} + {b^2} + 2ab \\

$

Using these results, we’ll get:

\[

\Rightarrow f(x + 3) = {(^4}{C_0}{x^4}{ + ^4}{C_1}{x^3}.3{ + ^4}{C_2}{x^2}{.3^2}{ + ^4}{C_3}x{.3^3}{ + ^4}{C_4}{.3^4}) - 12({x^3} + {3^3} + 3 \times x \times 3(x + 3)) \\

{\text{ }} + 17({x^2} + 9 + 6x) - 9(x + 3) + 7 \\

\Rightarrow f(x + 3) = ({x^4} + 12{x^3} + 54{x^2} + 108x + 81) - 12({x^3} + 9{x^2} + 27x + 27) + 17{x^2} + 153 + 102x - 9x - 27 + 7 \\

\Rightarrow f(x + 3) = {x^4} + 12{x^3} - 12{x^3} + 54{x^2} - 108{x^2} + 17{x^2} + 108x - 324x + 102x - 9x + 81 - 324 + 153 - 27 + 7 \\

\Rightarrow f(x + 3) = {x^4} - 37{x^2} - 123x - 110 \\

\]

Therefore, the final value of $f(x + 3)$is \[{x^4} - 37{x^2} - 123x - 110\]

Note: For ${(x + 3)^4},$instead of using binomial expansion we can also use general expansion method as:

${(x + 3)^4} = {(x + 3)^2}.{(x + 3)^2},$

${(x + 3)^4} = ({x^2} + 9 + 6x).({x^2} + 9 + 6x)$and we can multiply it step by step. We’ll get the same result.

The given function in the question is $f(x) = {x^4} - 12{x^3} + 17{x^2} - 9x + 7$

For $f(x + 3),$we will replace $x$with $x + 3,$we’ll get:

$ \Rightarrow f(x + 3) = {(x + 3)^4} - 12{(x + 3)^3} + 17{(x + 3)^2} - 9(x + 3) + 7$

For ${(x + 3)^4},$we’ll use binomial expansion and for rest of the terms we use formulae as:

$

{(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) \\

{(a + b)^2} = {a^2} + {b^2} + 2ab \\

$

Using these results, we’ll get:

\[

\Rightarrow f(x + 3) = {(^4}{C_0}{x^4}{ + ^4}{C_1}{x^3}.3{ + ^4}{C_2}{x^2}{.3^2}{ + ^4}{C_3}x{.3^3}{ + ^4}{C_4}{.3^4}) - 12({x^3} + {3^3} + 3 \times x \times 3(x + 3)) \\

{\text{ }} + 17({x^2} + 9 + 6x) - 9(x + 3) + 7 \\

\Rightarrow f(x + 3) = ({x^4} + 12{x^3} + 54{x^2} + 108x + 81) - 12({x^3} + 9{x^2} + 27x + 27) + 17{x^2} + 153 + 102x - 9x - 27 + 7 \\

\Rightarrow f(x + 3) = {x^4} + 12{x^3} - 12{x^3} + 54{x^2} - 108{x^2} + 17{x^2} + 108x - 324x + 102x - 9x + 81 - 324 + 153 - 27 + 7 \\

\Rightarrow f(x + 3) = {x^4} - 37{x^2} - 123x - 110 \\

\]

Therefore, the final value of $f(x + 3)$is \[{x^4} - 37{x^2} - 123x - 110\]

Note: For ${(x + 3)^4},$instead of using binomial expansion we can also use general expansion method as:

${(x + 3)^4} = {(x + 3)^2}.{(x + 3)^2},$

${(x + 3)^4} = ({x^2} + 9 + 6x).({x^2} + 9 + 6x)$and we can multiply it step by step. We’ll get the same result.

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