X=9 mark as brainlest if this helps ;)
        
                    
             
        
        
        
Answer: 3a
2
 (5a+2b)−5b
2
 (5a+2b)
2 Factor out the common term 5a+2b5a+2b.
(5a+2b)(3{a}^{2}-5{b}^{2})(5a+2b)(3a
2
 −5b
2
 )
Step-by-step explanation:
 
        
             
        
        
        
The correct answer is B :)
        
             
        
        
        
 The answer would be D) dilated by a scale factor of 3
Proof : 
A is at (-2, 1), A' is at (-6, 3) Try multiplying -2 by 3 = 6, 1 x 3 = 3. 
I hope this makes sense and helps.
 
        
             
        
        
        
According to the fundamental theorem of algebra, a polynomial has the
number of zeros equivalent to the degree of the polynomial.
A quadratic equation is of degree 2 and hence has 2 zeros.
A complex zero is of the form 

 which represents two zeros: a + bi and a - bi.
Thus,
since a quadratic equation has only 2 zeros, a quadratic equation with a
complex zero already have two zeros and hence will not have any other
zero.