The problem asks to factorize the quadratic expression $2x^2 + 9x + 4$.

AlgebraQuadratic EquationsFactorizationAlgebraic Manipulation
2025/5/10

1. Problem Description

The problem asks to factorize the quadratic expression 2x2+9x+42x^2 + 9x + 4.

2. Solution Steps

We need to factorize the quadratic expression 2x2+9x+42x^2 + 9x + 4. We are looking for two binomials of the form (ax+b)(cx+d)(ax + b)(cx + d) such that ac=2ac = 2, bd=4bd = 4, and ad+bc=9ad + bc = 9.
We can try different combinations of factors for 2 and

4. Since $2$ is prime, the only factors are $1$ and $2$. The factors of $4$ are $1$ and $4$ or $2$ and $2$.

Let's try a=2a = 2 and c=1c = 1. Then we have (2x+b)(x+d)(2x + b)(x + d). We want bd=4bd = 4 and 2d+b=92d + b = 9.
If b=1b = 1 and d=4d = 4, then 2d+b=2(4)+1=8+1=92d + b = 2(4) + 1 = 8 + 1 = 9. This works.
Thus, the factors are (2x+1)(x+4)(2x + 1)(x + 4).
Alternatively, we can use the "たすき掛け" (cross multiplication) method.
We look for two numbers that multiply to 24=82 \cdot 4 = 8 and add to 99. These numbers are 88 and 11.
Then we rewrite the middle term:
2x2+9x+4=2x2+8x+x+42x^2 + 9x + 4 = 2x^2 + 8x + x + 4
=2x(x+4)+1(x+4)= 2x(x + 4) + 1(x + 4)
=(2x+1)(x+4)= (2x + 1)(x + 4)

3. Final Answer

The factored form of the expression is (2x+1)(x+4)(2x + 1)(x + 4).