To find two trinomials \( f(x) \) and \( g(x) \) such that their sum equals \( 3x - 2 \), we can express them in the form:
\[ f(x) = ax^2 + bx + c \] \[ g(x) = dx^2 + ex + f \]
The sum of these two trinomials can be expressed as:
\[ f(x) + g(x) = (a + d)x^2 + (b + e)x + (c + f) \]
We want this to equal \( 3x - 2 \). For this equality to hold, the coefficients of the like terms must be equal. Thus, we have:
- \( a + d = 0 \) (no \( x^2 \) term in \( 3x - 2 \))
- \( b + e = 3 \)
- \( c + f = -2 \)
From this system, we can choose values for \( a, b, c, d, e, \) and \( f \). Let's choose:
- Let \( a = 0 \) and \( d = 0 \) (which means neither trinomial has an \( x^2 \) term).
- For \( b \), we can choose \( b = 3 \) and then set \( e = 0 \) (so \( g(x) \) has no \( x \) term).
- Finally, let \( c = -2 \) and \( f = 0 \) (so \( g(x) \) has no constant term).
This gives us:
\[ f(x) = 0x^2 + 3x - 2 = 3x - 2 \] \[ g(x) = 0x^2 + 0x + 0 = 0 \]
However, this doesn't give us two distinct trinomials. Let's revise this to ensure we have two distinct ones.
Instead, let:
- \( f(x) = x^2 + 2x - 2 \)
- \( g(x) = -x^2 + x \)
Now, let's check their sum:
\[ f(x) + g(x) = (x^2 - x^2) + (2x + x) + (-2) = 0x^2 + 3x - 2 \]
This indeed equals \( 3x - 2 \).
Thus, two trinomials \( f(x) \) and \( g(x) \) such that their sum is \( 3x - 2 \) are:
\[ f(x) = x^2 + 2x - 2 \] \[ g(x) = -x^2 + x \]