Find two functions f and g such that (f · g)(x) = h(x). (There are many correct answers.) h(x)=(2x+1)2

Question

crocolylec · Accepted Answer

Step 1  There are many possible answers.  Step 2  f(x)=x2g(x)=2x+1  f(x)=xg(x)=(2x+1)2  f(x)=x3g(x)=(2x+1)23  f(x)=x+1(2x+1)2−1

user_27qwe · Accepted Answer

Step 1:To solve the equation (f·g)(x)=h(x)=(2x+1)2, we need to find two functions, f(x) and g(x), such that their product yields the given function h(x).Let&#039;s express the functions f(x) and g(x) in terms of their respective components:f(x)=a1xm1+a0g(x)=b1xn1+b0where a1, a0, b1, and b0 are constants, and m1 and n1 are exponents.Step 2:Now, let&#039;s substitute these expressions into the equation and solve for the constants and exponents:(f·g)(x)=(a1xm1+a0)(b1xn1+b0)Expanding the above equation:(f·g)(x)=a1b1xm1+n1+a1b0xm1+a0b1xn1+a0b0We need this expression to be equal to (2x+1)2. Comparing the coefficients of the powers of x on both sides of the equation, we can determine the values of the constants and exponents.Comparing the coefficients of x2:a1b1=4Comparing the coefficients of x1:a1b0+a0b1=4Comparing the coefficients of x0:a0b0=1Step 3:Now, we can solve these equations to find the values of the constants:From the equation a0b0=1, we can set a0=1 and b0=1.Substituting these values into the equation a1b0+a0b1=4:a1+a0b1=4a1+b1=4We can set a1=3 and b1=1, or a1=2 and b1=2, or any other combination that satisfies the equation.Therefore, one possible solution is:f(x)=3x+1g(x)=x+1

karton · Accepted Answer

Result:f(x)=x2+2x and g(x)=4x2+2xSolution:Let&#039;s express f(x) and g(x) as polynomial functions and solve for their coefficients.Let f(x)=ax2+bx+c and g(x)=dx2+ex+f, where a, b, c, d, e, and f are the coefficients we need to determine.The product of f(x) and g(x) is given by:(f·g)(x)=(ax2+bx+c)(dx2+ex+f)Expanding the above expression, we have:(f·g)(x)=adx4+(ae+bd)x3+(af+be+cd)x2+(bf+ce)x+cfComparing this with the given function h(x)=(2x+1)2=4x2+4x+1, we can equate the coefficients:ad=4ae+bd=4af+be+cd=1bf+ce=0cf=0Solving this system of equations, we can find multiple solutions. Let&#039;s consider one possible solution:Let a=1, b=2, c=0, d=4, e=2, and f=0.Substituting these values into the expressions for f(x) and g(x), we get:f(x)=x2+2xg(x)=4x2+2xTherefore, one solution for f and g is f(x)=x2+2x and g(x)=4x2+2x.

alenahelenash · Accepted Answer

To find two functions f and g such that (f·g)(x)=h(x)=(2x+1)2, we can approach the problem by factoring the given expression. Let&#039;s express h(x) as h(x)=(2x+1)(2x+1).Now, we can express f(x) and g(x) in terms of the factors of h(x) as follows:f(x)=2x+1 and g(x)=2x+1.To verify that this is a valid solution, let&#039;s compute the product (f·g)(x):(f·g)(x)=f(x)·g(x)=(2x+1)·(2x+1)=(2x+1)2.Thus, the functions f(x)=2x+1 and g(x)=2x+1 satisfy the given condition.

Find two functions f and g such that (f · g)(x) = h(x). (There are many correct

Answered question

Answer & Explanation

New Questions in Algebra I