This is in integers. q: evaluate (++)

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rayraye

01.09.2022

This question is, in other words, asking when the graph of $y = x^3 - 7x^2 + 36$ is negative (below the x-axis).we have to factor the leftside first to find the zeros. the zeros of the function $y = x^3 - 7x^2 + 36$ will be where the graph will cross the x-axis.rational root theorem tells us that possible factors of this polynomial are possible factors of the constant term.factors of 36: $\pm 1, \pm 2, \pm 3, \pm4, /tex]we use rational root theorem with remainder theorem, which tells us that putting a value [tex]a$ in a function gets us the remainder of dividing that function by $x-a$ test those factors to find a value that makes the function zero so that we can find the right binomial to divide by to get a remainder of zero. $y(1) = (1)^3 - 7(1)^2 + 36 = 30$ : no $y(-1) = (-1)^3 - 7(-1)^2 + 36 = 28$ : no $y(2) = (2)^3 - 7(2)^2 + 36 = 16$ : no $y(-2) = (-2)^3 - 7(-2)^2 + 36 = 0$ : yessince putting in $x=-2$ makes the function zero, then dividing $x^3 - 7x^2 + 36$ by $x+2$ will have remainder of zero; this tells us that $x+2$ is factor.division is performed as attachment, with both synthetic and long division as options. either way, quotient of $x^2-9x+18$ . the division statement for this is $\dfrac{x^3 - 7x^2 + 36}{x+2} = x^2-9x+18$ multiplying both sides by $x+2$ : $x^3 - 7x^2 + 36 = (x+2)(x^2-9x+18)$ therefore $\begin{aligned} x^3 - 7x^2 + 36 & \ \textless \ 0 \\ (x+2)(x^2-9x+18) & \ \textless \ 0 \end{aligned}$ $x^2-9x+18$ is factorable using normal methods. two numbers that multiply to get +18 and add to get -9: -6 and -3 $\begin{aligned}(x+2)(x^2-9x+18)\ \textless \ 0 \\ (x+2)(x-3)(x-6)\ \textless \ 0 \end{aligned}$ you can either graph this polynomial or use region testing.we have the following regions to test: region 1: $x \ \textless \ -2$ region 2: $-2 \ \textless \ x \ \textless \ 3$ region 3: $3\ \textless \ x\ \textless \ 6$ region 4: $x \ \textgreater \ 6$ we use the zeros as region boundaries because the zeros are the only place a polynomial function can cross the x-axis, which determines if the result is positive or negative. testing a value in a region formed by the zeros determines if the result is positive or negative for that entire region.pick a value for each region to test if the results is negative or positive. if it is negative, then the region does satisfy $x^3 - 7x^2 + 36\ \textless \ 0$ .region 1: possible value to test in $x \ \textless \ -2$ is -3: $(-3)^3 - 7(-3)^2 + 36 = -54$ so $x\ \textless \ -2$ is a solution.region 2: possible value to test in $-2 \ \textless \ x \ \textless \ 3$ is 0 $(0)^3 - 7(0)^2 + 36 = 36$ this result is positive, so this region is not a solution.region 3: possible value to test in $3\ \textless \ x\ \textless \ 6$ is 4 $(4)^3 - 7(4)^2 + 36 = -12$ so $3\ \textless \ x\ \textless \ 6$ is a solution.region 4: possible value to test in $x \ \textgreater \ 6$ is 7. $(7)^3 - 7(7)^2 + 36 = 36$ this result is positive, so this region is not a solution.so the values that satify are: $x \ \textless \ -2,\ 3 \ \textless \ x \ \textless \ 6$

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