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  • The factorization of 4x^{2}+8x+3 is (A) (x+1)(x+3) (B) (2x+1)(2x+3) (C) (2x+2)(2x+5) (D) (2x-1)(2x-3)

The factorization of 4x^{2}+8x+3  is

(A) (x+1)(x+3)       

(B) (2x+1)(2x+3)    

(C) (2x+2)(2x+5)   

(D) (2x-1)(2x-3)

Answers (1)

Let us factorize the given polynomial 4x^{2}+8x+3

To factorize ax2 + bx + c, we have to distribute “bx” into “px” and “qx” such that

p + q = b and p.q = a.c

So, 4x^{2}+8x+3=0

can be written as

4x^{2}+6x+2x+3=0      {6+2 = 8; (6)(2) = (4)(3)}

\\2x(2x+3)+1(2x+3)=0\\ (2x+3)(2x+1)=0\\

Hence option B is correct

Posted by

infoexpert24

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