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Chapter 2: Problem 131

Simplify: \(2(x+h)^{2}+3(x+h)+5-\left(2 x^{2}+3 x+5\right)\)

Short Answer

Expert verified
The simplified form of the expression is \(4hx+2h^2+3h-5\)

Step by step solution

01

Expansion of the expression

Expand \(2(x+h)^{2}+3(x+h)\) using the formula \((a+b)^2 = a^2 + 2ab + b^2\). This gives us \(2(x^2+2xh+h^2)+3x+3h\), which can be further simplified to \(2x^2+4hx+2h^2+3x+3h\).
02

Subtract the second expression

Subtract \(2 x^{2}+3 x+5\) from the expanded expression. We have: \(2x^2+4hx+2h^2+3x+3h - (2x^2+3x+5)\). Distribute the negative sign into the parenthesis. The expression becomes \(2x^2+4hx+2h^2+3x+3h - 2x^2 - 3x - 5\).
03

Grouping like terms and simplify

Now, let's group the common terms and simplify the expression. The terms \(2x^2\), \(4hx\), \(2h^2\), \(3x\), \(3h\) are matched with \(-2x^2\), \(-3x\), \(-5\) respectively. This leaves us with the expression \(4hx+2h^2+3h-5\).

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Most popular questions from this chapter

Does \((x-3)^{2}+(y-5)^{2}=0\) represent the equation of a circle? If not, describe the graph of this equation.

Suppose that \(h(x)=\frac{f(x)}{g(x)} .\) The function \(f\) can be even, odd, or neither. The same is true for the function \(g .\) a. Under what conditions is \(h\) definitely an even function? b. Under what conditions is \(h\) definitely an odd function?

give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation's domain and range. $$ (x-2)^{2}+(y-3)^{2}=16 $$

write the standard form of the equation of the circle with the given center and radius. $$ \text { Center }(-1,4), r=2 $$

Define a piecewise function on the intervals \((-\infty, 2],(2,5)\) and \([5, \infty)\) that does not "jump" at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.
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