Question

Simplify: 4k – 2 + 3k2 – 5k + 6
(a) k2 + k +4
(c) 3k2 - 9k +4
(b) 3k2 – k +4
(d) 3k2 - 9k - 4
please explain also​

Answer 1

Answer:

b is correct

Step-by-step explanation:

Plz Mark as Brilliant

Answer 2

Question:-

Simplify: 4k – 2 + 3k2 – 5k + 6

(a) k2 + k +4

(c) 3k2 - 9k +4

(b) 3k2 – k +4

(d) 3k2 - 9k - 4

Answer:-

$\sf \: 6 - 5k {}^{2}$

Step by Step Solution:-

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "k2" was replaced by "k^2". 3 more similar replacement(s).

STEP 1:

Equation at the end of step 1

((((4•(k2))+(3•(k2)))-(7•(k2)))-5k2)+6

STEP 3:

Equation at the end of step 3:

((((4•(k2))+3k2)-7k2)-5k2)+6

STEP 4:

Equation at the end of step 4:

(((22k2 + 3k2) - 7k2) - 5k2) + 6

STEP 5:

Trying to factor as a Difference of Squares:

5.1 Factoring: 6-5k2

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof :

$\sf(A+B) • (A-B) = \\ A {}^{2}   \sf- AB + BA - B {}^{2}  = \\ A {} \sf^{2} - AB + AB - B {}^{2}  = \\ </p><p> \sf \: A {}^{2}  - B {}^{2}$

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 6 is not a square !!

Ruling : Binomial can not be factored as the

difference of two perfect squares

Final result :

$\sf \: 6 - 5k {}^{2}$