Question #13537

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Question #13537

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Answer 1 · 2017-05-03T19:09:12

a) $g (5 x + 2) = 75 x^{2} + 35 x + 4$ $g(5x+2) = 75x^2+35x+4$
b) $f (x) - h (5) = 2 x - 125$ $f(x) - h(5) = 2x-125$
c) $f (x) [h (x) - g (x)] = 2 x^{4} - 11 x^{3} + 23 x^{2} - 24 x + 10$ $f(x)[h(x)-g(x)] = 2x^4-11x^3+23x^2-24x+10$

Explanation:

Given:

$f (x) = 2 x - 5$ $f(x) = 2x-5$
$g (x) = 3 x^{2} - 5 x + 2$ $g(x) = 3x^2-5x+2$
$h (x) = x^{3} - x$ $h(x) = x^3-x$

a) In order to find $g (5 x + 2)$ $g(5x+2)$ we must plug in for every value of $x$ $x$ in $g (x)$ $g(x)$ the value $(5 x + 2)$ $(5x+2)$ .

$g (x) = 3 x^{2} - 5 x + 2$ $g(x) = 3x^2-5x+2$
$g (5 x + 2) = 3 {(5 x + 2)}^{2} - 5 (5 x + 2) + 2$ $g(5x+2) = 3(5x+2)^2-5(5x+2)+2$

Now we simplify the expression on the right hand side.

$= 3 (5 x + 2) (5 x + 2) - 5 (5 x + 2) + 2$ $= 3(5x+2)(5x+2)-5(5x+2)+2$

We FOIL $(5 x + 2) (5 x + 2)$ $(color(red)(5x)color(blue)(+2))(color(green)(5x)color(purple)(+2))$ and distribute $(- 5)$ $(-5)$ to $(5 x + 2)$ $(5x+2)$

$= 3 [(5 x) (5 x) + (5 x) (+ 2) + (2) (5 x) + (+ 2) (+ 2)] - 25 x - 10 + 2$ $= 3[(color(red)(5x))(color(green)(5x))+(color(red)(5x))(color(purple)(+2))+(color(blue)(2))(color(green)(5x))+(color(blue)(+2))(color(purple)(+2))]-25x-10+2$

$= 3 [25 x^{2} + 10 x + 10 x) + 4] - 25 x - 8$ $= 3[25x^2+10x+10x)+4]-25x-8$

$= 75 x^{2} + 60 x + 12 - 25 x - 8$ $= 75x^2+60x+12-25x-8$

$g (5 x + 2) = 75 x^{2} + 35 x + 4$ $g(5x+2) = 75x^2+35x+4$

b) For this part, we again substitute the given value of $5$ $5$ into $h (x)$ $h(x)$

$f (x) = 2 x - 5$ $f(x) = 2x-5$
$h (x) = x^{3} - x$ $h(x) = x^3-x$

$f (x) - h (5) = 2 x - 5 - [{(5)}^{3} - (5)]$ $f(x) - h(5) = 2x-5 - [(5)^3-(5)]$

$= 2 x - 5 - 125 + 5$ $= 2x-5 - 125 + 5$

$f (x) - h (5) = 2 x - 125$ $f(x) - h(5) = 2x-125$

c) For this part, we need only to perform arithmetic on the given functions:

$f (x) = 2 x - 5$ $f(x) = 2x-5$
$g (x) = 3 x^{2} - 5 x + 2$ $g(x) = 3x^2-5x+2$
$h (x) = x^{3} - x$ $h(x) = x^3-x$

$f (x) [h (x) - g (x)] = (2 x - 5) [x^{3} - x - (3 x^{2} - 5 x + 2)]$ $f(x)[h(x)-g(x)] = (2x-5)[x^3-x-(3x^2-5x+2)]$

Simplifying the expression in the brackets, we get
$= (2 x - 5) [x^{3} - x - 3 x^{2} + 5 x - 2]$ $= (2x-5)[x^3-x-3x^2+5x-2]$
$= (2 x - 5) [x^{3} - 3 x^{2} + 4 x - 2]$ $= (2x-5)[x^3-3x^2+4x-2]$

Now we need to expand and multiply $(2 x - 5)$ $(2x-5)$ by $[x^{3} - 3 x^{2} + 4 x - 2]$ $[x^3-3x^2+4x-2]$

$= (2 x - 5) [x^{3} - 3 x^{2} + 4 x - 2]$ $=(color(red)(2x)color(blue)(-5))[x^3-3x^2+4x-2]$

We can distribute the terms in $(2 x - 5)$ $(2x-5)$ and multiply each of them by $[x^{3} - 3 x^{2} + 4 x - 2]$ $[x^3-3x^2+4x-2]$

$= (2 x) [x^{3} - 3 x^{2} + 4 x - 2] + (- 5) [x^{3} - 3 x^{2} + 4 x - 2]$ $=(color(red)(2x))[x^3-3x^2+4x-2]+(color(blue)(-5))[x^3-3x^2+4x-2]$

This gives:

= $2 x^{4} - 6 x^{3} + 8 x^{2} - 4 x$ $color(red)(2x^4-6x^3+8x^2-4x)$
$- 5 x^{3} + 15 x^{2} - 20 x + 10$ $color(blue)(-5x^3+15x^2-20x+10)$

Combining like terms, gives:

$= 2 x^{4} + (- 6 - 5) x^{3} + (8 + 15) x^{2} + (- 4 - 20) x + 10$ $=color(red)(2)x^4 + (color(red)(-6)color(blue)(-5))x^3 + (color(red)(8)color(blue)(+15))x^2 + (color(red)(-4)color(blue)(-20))x+color(blue)(10)$

$f (x) [h (x) - g (x)] = 2 x^{4} - 11 x^{3} + 23 x^{2} - 24 x + 10$ $f(x)[h(x)-g(x)] = 2x^4-11x^3+23x^2-24x+10$

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Question #13537

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