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zvonat [6]
3 years ago
5

(z-2) (z+6) Multiplying polynomials

Mathematics
2 answers:
Likurg_2 [28]3 years ago
8 0
x^{2} +6z-2z+12 = x^{2} +4z+12

jolli1 [7]3 years ago
3 0


z2-z-6=0

Two solutions were found :

z = 3
z = -2
Reformatting the input :

Changes made to your input should not affect the solution:

(1): "z2" was replaced by "z^2".

Step by step solution :

Step 1 :

Trying to factor by splitting the middle term

1.1 Factoring z2-z-6

The first term is, z2 its coefficient is 1 .
The middle term is, -z its coefficient is -1 .
The last term, "the constant", is -6

Step-1 : Multiply the coefficient of the first term by the constant 1 • -6 = -6

Step-2 : Find two factors of -6 whose sum equals the coefficient of the middle term, which is -1 .

-6 + 1 = -5
-3 + 2 = -1 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 2
z2 - 3z + 2z - 6

Step-4 : Add up the first 2 terms, pulling out like factors :
z • (z-3)
Add up the last 2 terms, pulling out common factors :
2 • (z-3)
Step-5 : Add up the four terms of step 4 :
(z+2) • (z-3)
Which is the desired factorization

Equation at the end of step 1 :

(z + 2) • (z - 3) = 0
Step 2 :

Theory - Roots of a product :

2.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

2.2 Solve : z+2 = 0

Subtract 2 from both sides of the equation :
z = -2

Solving a Single Variable Equation :

2.3 Solve : z-3 = 0

Add 3 to both sides of the equation :
z = 3

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A transformation T : (x, y) → (x + 3, y + 1). Find the preimage of the point (4, 3) under the given transformation. (7, 4) (1, 2
motikmotik

Answer:

(1, 2)

Step-by-step explanation:

Remember that the final shape and position of a figure after a transformation is called the image, and the original shape and position of the figure is the pre-image.

In our case, our figure is just a point. We know that after the transformation T : (x, y) → (x + 3, y + 1), our image has coordinates (4, 3).

The transformation rule T : (x, y) → (x + 3, y + 1) means that we add 3 to the x-coordinate and add 1 to the y-coordinate of our pre-image. Now to find the pre-image of our point, we just need to reverse those operations; in other words, we will subtract 3 from the x-coordinate and subtract 1 from the y-coordinate.

So, our rule to find the pre-image of the point (4, 3) is:

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We know that the x-coordinate of our image is 4 and its y-coordinate is 3.

Replacing values:

                (4 - 3, 3 - 1)

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We can conclude that our pre-image is the point (1, 2).

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