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  1. #1
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    3 questions

    Solve the following inequality. Write the answer in interval notation.
    Note: If the answer includes more than one interval write the intervals separated by the "union" symbol, U. If needed enter as infinity and as -infinity .




    Find the quotient and remainder using long division for






    Simplify the expression
    and give your answer in the form of

    Your answer for the function is :
    Your answer for the function is :
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  2. #2
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    Quote Originally Posted by wannous View Post
    Solve the following inequality. Write the answer in interval notation.
    Note: If the answer includes more than one interval write the intervals separated by the "union" symbol, U. If needed enter as infinity and as -infinity .




    Find the quotient and remainder using long division for








    Simplify the expression



    and give your answer in the form of



    Your answer for the function is :


    Your answer for the function is :
    For Q.3 note that

    $\displaystyle 7x^3 - 8x^2 - 12x = x(x-2)(7x+3)$

    and

    $\displaystyle 6x^2 - 16x + 8 = 2(x - 2)(3x - 2)$.


    So $\displaystyle \frac{7x^3 - 8x^2 - 12x}{6x^2 - 16x + 8} = \frac{x(x-2)(7x+3)}{2(x - 2)(3x - 2)}$

    $\displaystyle = \frac{x(7x + 3)}{2(3x - 2)}$.


    Therefore $\displaystyle f(x) = x(7x + 3)$ and $\displaystyle g(x) = 2(3x - 2)$.
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  3. #3
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    Hello wannous
    Quote Originally Posted by wannous View Post
    Solve the following inequality. Write the answer in interval notation.
    Note: If the answer includes more than one interval write the intervals separated by the "union" symbol, U. If needed enter as infinity and as -infinity .

    To solve this type of inequality, find the values of $\displaystyle x$ that make each factor zero, and then look at their signs as $\displaystyle x$ moves along the number line from left to right, through each of these 'zero values' in turn.

    Now the 'zeros' occurs when $\displaystyle x =4$ and $\displaystyle x=5$. So we look at what happens when:

    • $\displaystyle x<4$


    • $\displaystyle 4<x<5$


    • $\displaystyle 5<x$

    First, if $\displaystyle x<4,\, (x-5)$ is negative and $\displaystyle (x-4)$ is also negative; and $\displaystyle -^{ve}\times -^{ve} = +^{ve}$. So $\displaystyle (x-5)(x-4)>0$ in this range.

    Next, if $\displaystyle 4<x<5,\, (x-5)$ is $\displaystyle -^{ve}$ and $\displaystyle (x-4)$ is $\displaystyle +^{ve}$; $\displaystyle -^{ve}\times+^{ve} = -^{ve}$. So $\displaystyle (x-5)(x-4) < 0$ in this range.

    Finally, if $\displaystyle x>5$, we get $\displaystyle +^{ve}\times+^{ve}=+^{ve}$. So $\displaystyle (x-5)(x-4)>0$ in this range.

    So the values we want are $\displaystyle 4\le x \le 5$, or in interval notation $\displaystyle [4,5]$.


    Find the quotient and remainder using long division for

    See the attached image.

    So the quotient is
    $\displaystyle x^2-4x+1$ and the remainder is $\displaystyle -5$.

    Grandad


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