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Kitty [74]
1 year ago
15

X^4+2x^2−63 solve by factoring group

Mathematics
1 answer:
Harrizon [31]1 year ago
3 0

The polynomial equation be x⁴ + 2x² - 63 then the factoring group exists  ( x² - 7) (x² + 9).

<h3>What is meant by factorization?</h3>

Factorization is the process of dividing a large number into smaller numbers that, when multiplied together, yield the original number. Factorization occurs when you divide a number into its factors or divisors.

Let the given polynomial equation be  x⁴ + 2x² - 63

By factoring the above polynomial equation, we get

Here the coefficient of the first term, x⁴, is 1.

The coefficient of the middle term, 2x², is 2.

The final term, "the constant," is -63.

To find -63 factors whose sum equals the middle term's coefficient, which is 2.

-63 + 1 = -62

-21 + 3 = -18

-9 + 7 = -2

-7 + 9 = 2

Split the middle term of the polynomial using the two factors found in step 2 above, -7 and 9 .

x⁴ - 7x² + 9x² - 63.

= ( x² - 7) (x² + 9)

∴ x⁴ + 2x² - 63 = ( x² - 7) (x² + 9) .

The polynomial equation be x⁴ + 2x² - 63 then the factoring group exists (x² - 7) (x² + 9).

To learn more about factorization refer to :

brainly.com/question/25829061

#SPJ1

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Write each of the following expressions without using absolute value: |z−6|−|z−5|, if z&lt;5
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Answer:

The answer is <em>1</em>.

Step-by-step explanation:

Given the expression:

|z-6|-|z-5|,\ if\ z

To find:

The expression without absolute value.

Solution:

First of all, let us learn about the absolute value function:

y = f(x) = |x| =\left \{ {{x\ if\ x>0} \atop {-x\ if\ x

i.e. value is x if x is positive

value is -x if x is negative

Here the given expression contains two absolute value functions:

|z-6| and |z-5|

Using the definition of absolute value function as per above definition.

|z-5| =\left \{ {{(z-5)\ if\ z>5} \atop {-(z-5)\ if\ z

|z-6| =\left \{ {{(z-6)\ if\ z>6} \atop {-(z-6)\ if\ z

Now, it is given that z < 5 that means z will also be lesser than 6 i.e. z < 6

So, given expression |z-6|-|z-5|,\ if\ z will be equivalent to :

-(z-6) - (-(z-5))\\\Rightarrow -z+6 +z-5 = \bold{1}

So, the expression is equivalent to <em>1</em>.

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