The factorisation of 4 x^{2}+8 x+3 is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To factorise the quadratic expression $4 x^{2}+8 x+3$,we use the splitting the middle term method.We need to find two numbers whose product is $4

Vedclass · Accepted Answer

$(2 x+1)(2 x+3)$

Vedclass · Answer

(A) To factorise the quadratic expression $4 x^{2}+8 x+3$,we use the splitting the middle term method.We need to find two numbers whose product is $4 	imes 3 = 12$ and whose sum is $8$.These two numbers are $6$ and $2$,since $6 	imes 2 = 12$ and $6 + 2 = 8$.Now,rewrite the middle term $8x$ as $6x + 2x$:$4 x^{2}+6 x+2 x+3$Group the terms:$(4 x^{2}+6 x) + (2 x+3)$Factor out the common terms from each group:$2 x(2 x+3) + 1(2 x+3)$Finally,factor out the common binomial $(2 x+3)$:$(2 x+3)(2 x+1)$Thus,the correct option is $A$.

The factorisation of $4 x^{2}+8 x+3$ is

Similar Questions

Verify whether the following is True or False:
$-3$ is a zero of $x-3$.

Write the coefficient of $x^{2}$ in the following polynomial:
$3x^{3} - 8x^{2} + 14x - 5$

Using a suitable identity,evaluate the following:
$103^{3}$

Write the following cube in expanded form: $(2x + 7)^3$.

For the polynomial $\frac{x^{3}+2 x+1}{5}-\frac{7}{2} x^{2}-x^{6},$ write:
$(i)$ the degree of the polynomial
$(ii)$ the coefficient of $x^{3}$
$(iii)$ the coefficient of $x^{6}$
$(iv)$ the constant term

Vedclass Test Series

Exam Paper Generator

Online Exam Module