MATHEMATICS HIGH SCHOOL

Use Synthetic Division to Factor the following polynomials completely by given factors or roots and Find all the zeros; When you reach quadratic equation, performance regular factoring or Quadratic Formula. x^(5) -9x^(3) -x^(2) +9;(x-3),(x-1), as factors and -3 as a root

Answers

Answer 1
Answer:

Answer:

The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)

The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2

Step-by-step explanation:

(x^(5)+0-9x^(3) -x^(2)+0+9) ÷ (x - 3) =

(x^(4)-9x^(3)-x^(2)+0+9) ÷ (x - 3) =

x^(4)+3x^(3)+(-x^(2)+0+9) ÷ (x - 3) =

x^(4)+3x^(3)-x+(-3x+9) ÷ (x - 3) =

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

(x^(4)+3x^(3)+0-x-3) ÷ (x - 1) =

x³ + (4x³ + 0 - x - 3) ÷ (x - 1) =

x³ + 4x² + (4x² - x - 3) ÷ (x - 1) =

x³ + 4x² + 4x + (3x - 3) ÷ (x - 1) =

x³ + 4x² + 4x + 3

∵ -3 is a root ⇒ (x + 3) is a factor

(x³ + 4x² + 4x + 3) ÷ (x + 3) =

x² + (x² + 4x + 3) ÷ (x + 3) =

x² + x + (x + 3) ÷ (x + 3) =

x² + x + 1 ⇒ use the formula to find the factors of this quadratic

∵ a = 1 , b = 1 and c = 1

∴ x=\frac{-1+\sqrt{(1)^(2)-4(1)(1)} }{2(1)}=(-1+√(-3))/(2)=(-1+i√(3))/(2)

∴ x=(-1-i√(3) )/(2)

∴ The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)

The zeroes:

x - 3 = 0 ⇒ x = 3

x + 3 = 0 ⇒ x = -3

x - 1 = 0 ⇒ x = 1

x - (1 - i√3)/2 = 0 ⇒ x = (1 - i√3)/2

x - (1 + i√3)/2 = 0 ⇒ x = (1 + i√3)/2

The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2

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I’m guessing it’s -22

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Please help me simplify these radicals without decimals

Answers

6) √72= √3*25= √3*2*12= √3*3*2*2*2 =3*2√2 =6√2
(just break down the numbers until they are prime numbers. )
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Hi There

6) √72 = 6√2

7) √98 = 7√2

8) √8= 2√2

9) √12= 2√3

10) √175 = 5√7

11) √216= 6√6

12) √192= 8√√3

13) √343 = 7√7

14) √32= 4√2

15) √28= 2√7


Hint : The first thing you need to find is the largest perfect square witch will divide evenly into the number under your radical sign .


I hope that's help !

What is the product of 3a 5 and 2a2 4a – 2? 6a3 22a2 14a – 10 6a3 22a2 26a –10 18a3 10a2 14a – 10 28a3 14a – 10

Answers

The product of  3a + 5 and 2a2 + 4a – 2 is \rm 6a^3+22a^2+14a-10\n, the correctoption is A.

What is the product of two algebraic expressions?

When two algebraic expressions are multiplied, the result is called the product, and the two expressions making up the product are called factors and multiplicands.

The multiplication of two algebraic equations can be expressed as a × b.

The product of 3a + 5 and 2a2 + 4a – 2 is determined in the following steps given below.

\rm =(3a + 5) *( 2a^2 + 4a -2)\n\n= 3a ( 2a^2 + 4a -2) +5( 2a^2 + 4a -2)\n\n =6a^3+12a^2-6a+10a^2+20a-10\n\n=6a^3+22a^2+14a-10\n

Hence, the product of  3a + 5 and 2a2 + 4a – 2 is \rm 6a^3+22a^2+14a-10\n.

To know more about multiplication click the link given below.

brainly.com/question/14059007

Answer:

6a3 + 22a2 + 14a - 10

Step-by-step explanation:

(3a + 5) x (2a^2 + 4a - 2)

= 3a(2a^2 + 4a - 2) + 5(2a^2 + 4a - 2)

= 6a^3 + 12a^2 - 6a + 10a^2 + 20a - 10

= 6a^3 + 22a^2 + 14a - 10

PLEASE HELP QUICK I WILL GIVE 67 POINTS!!!!! PLUS BRAINLIEST
SOLVE ALL OF IT :)

Answers

Step-by-step explanation:

Provided here is a photo.
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