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GuDViN [60]
3 years ago
15

12x^4 − 20x^2+ 32x

Mathematics
1 answer:
gulaghasi [49]3 years ago
6 0

1) common factors of the variable is x

2) common factors of the constants is 4

3) common factors of the expression is 4x

<em><u>Solution:</u></em>

Given that,

12x^4 - 20x^2 + 32x

<em><u>1. Explain how you find the common factors of the variable? </u></em>

The variables are:

x^4\\\\x^2\\\\x

Find the factors:

x^4 = x \times x \times x \times x\\\\x^2 = x \times x\\\\x

Here, "x" is present in all variables

x is the greatest common factor

Thus common factors of the variable is x

<em><u>2, Explain how you find the common factors of the constants?</u></em>

The constants are 12, 20 and 32

Find the greatest common factor

List down factors of 12, 20 and 32

The factors of 12 are: 1, 2, 3, 4, 6, 12

The factors of 20 are: 1, 2, 4, 5, 10, 20

The factors of 32 are: 1, 2, 4, 8, 16, 32

Then the greatest common factor is 4

The common factors of the constants is 4

<em><u>3. What are the common factors of the expression?</u></em>

Given is:

12x^4 - 20x^2 + 32x

Factor out x and 4

4x(3x^3 - 5x + 8)

The common factors of the expression is 4x

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based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

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\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)


\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}


now, we can plug those values on A = (1/2)bh,


\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5

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