Answer:
see explanation
Step-by-step explanation:
(a)
Given
2k - 6k² + 4k³ ← factor out 2k from each term
= 2k(1 - 3k + 2k²)
To factor the quadratic
Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)
The factors are - 1 and - 2
Use these factors to split the k- term
1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )
1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term
= (1 - k)(1 - 2k)
1 - 3k + 2k² = (1 - k)(1 - 2k) and
2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)
(b)
Given
2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )
= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term
= (x - 2y)(2a + 3b)
Given:
Polynomial is 
.
To find:
The sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form.
Solution:
The sum of given polynomial and the square of the binomial (x-8) is
![[\because (a-b)^2=a^2-2ab+b^2]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5D)
On combining like terms, we get
Therefore, the sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form is 
.
6x+19.98=200
200-19.98=6x
X= 200-19.98/6
= solve for it.
Can you please explain what question you have?
