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2 years ago

Factor (3c + 2)^2 - (3c - 2)^2

Mathematics

1 answer:

kodGreya [7K]2 years ago

6 0

Hi there! Your answe is 24c

Please see an explanation below for a clear understanding to your problem.

Any questions about my explanation and answer can be asked through comments! :)

Step-by-step explanation:

$\large \boxed{ {x}^{2} - {y}^{2} = (x + y)( x - y)}$

The formula above is what we will be using. This is called "Difference of Two Squares".

Notice how the these expressions are similar to the formula's structure.

$\large{ {(3c + 2)}^{2} - {(3c - 2)}^{2} }$

Notice how there are minus sign, squared expression which we can factor by using the formula above. Therefore -

$\large{ {x}^{2} = {(3c + 2)}^{2} } \\ \large{x = 3c + 2}$

Therefore, our y would be 3c-2.

Then proceed with the formula.

$\large{ {(3c + 2)}^{2} - {(3c - 2)}^{2} = (3c + 2 + 3c - 2)(3c + 2 - (3c - 2))}$

Simplify.

$\large{ (6c)(3c + 2 - 3c + 2)} \\ 6c(4) \\ 24c$

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Find f(a), f(a+h), and 71. f(x) = 7x - 3 f(a+h)-f(a) h if h = 0. 72. f(x) = 5x²

Leni [432]

Answer:

$71. \ \ \ f(a) \ = \ 7a \ - \ 3; \ f(a+h) \ = \ 7a \ + \ 7h \ - \ 3; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 7$

$72. \ \ \ f(a) \ = \ 5a^{2}; \ f(a+h) \ = \ {5a}^{2} \ + \ 10ah \ + \ {5h}^{2}; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 10a \ + \ 5h$

Step-by-step explanation:

In single-variable calculus, the difference quotient is the expression

$\displaystyle\frac{f(x+h) \ - \ f(x)}{h}$ ,

which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).

This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.

$m \ \ = \ \ \displaystyle\frac{\Delta y}{\Delta x} \ \ = \ \ \displaystyle\frac{rise}{run}$ .

Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as h approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.

Therefore,

$71. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{(7a \ + \ 7h \ - \ 3) \ - \ (7a \ - \ 3)}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{7h}{h} \\ \\ \-\hspace{4.25cm} = \ \ 7$

$72. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{{5(a \ + \ h)}^{2} \ - \ {5(a)}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{{5a}^{2} \ + \ 10ah \ + \ {5h}^{2} \ - \ {5a}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{h(10a \ + \ 5h)}{h} \\ \\ \-\hspace{4.25cm} = \ \ 10a \ + \ 5h$

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Please help me. Factor. 5x^+23x+26

Alexus [3.1K]

20,897,614 solved | 1,346 online

Skip Ad 
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Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

5*x^2-23*x-(-26)=0

Step by step solution :Skip Ad
 Step 1 :Equation at the end of step 1 : (5x2 - 23x) - -26 = 0 Step 2 :Trying to factor by splitting the middle term

2.1 Factoring 5x2-23x+26

The first term is, 5x2 its coefficient is 5 .
The middle term is, -23x its coefficient is -23 .
The last term, "the constant", is +26 

Step-1 : Multiply the coefficient of the first term by the constant 5 • 26 = 130

Step-2 : Find two factors of 130 whose sum equals the coefficient of the middle term, which is -23 .

-130 + -1 = -131 
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