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Nady [450]
3 years ago
5

The polynomial 6x²+x–15 has a factor of 2x–3. What is the other factor?

Mathematics
1 answer:
spayn [35]3 years ago
3 0

Step-by-step explanation:

Method 1

Do the factor like

6x

6 \times  ^{2}  +  \times  - 15 \\  = 6 \times  ^{2}  - 9 \times  + 10 \times  - 15 \\  = 3 \times (2 \times  - 3) + 5(2 \times  - 3) \\  = (2 \times  - 3)(3 \times  + 5)

So the other factor is 3x+5

Method 2 We can divise 6x^2+x-15 by 2x-3 and we find the quotient 3x+5

Hope you understand

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Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are  equal, thus

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JH=\sqrt{(5-0)^{2}+(1-1)^{2}}=5 and

GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5

Now, mid point of JH=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

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Mid point of GI=(\frac{5}{2},1)

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

(GI)^{2}=(IJ)^{2}+(JG)^{2}                             (1)

Now, JI=\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20} and GJ=\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}

Substituting these values in (1), we get

5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }

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Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5  and HJ=\sqrt{(0-5)^2+(1-1)^2}=5 are equal, thus, GHIJ is a rectangle.

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