Answer :
Let's solve the problem by factorizing the given quadratic expression step by step.
Given the quadratic expression:
[tex]\[ 3x^2 + 26x + 51 \][/tex]
We want to factorize this expression and use the factors to find two numbers which, when multiplied, approximate or equal [tex]\( 32651 \)[/tex] when [tex]\( x = 1 \)[/tex].
### Step 1: Factorize the quadratic expression
First, recognize that factorizing a quadratic expression of the form [tex]\( ax^2 + bx + c \)[/tex] typically yields a product of two binomial expressions:
[tex]\[ 3x^2 + 26x + 51 = (Ax + B)(Cx + D) \][/tex]
We need to determine [tex]\(A\)[/tex], [tex]\(B\)[/tex], [tex]\(C\)[/tex], and [tex]\(D\)[/tex] such that:
[tex]\[ AC = 3 \quad (the\; coefficient\; of\; x^2) \][/tex]
[tex]\[ AD + BC = 26 \quad (the\; coefficient\; of\; x) \][/tex]
[tex]\[ BD = 51 \quad (the\; constant\; term) \][/tex]
Through the factorization process, we find that the quadratic can be factorized into:
[tex]\[ (3x + 17)(x + 3) \][/tex]
### Step 2: Substitute [tex]\( x = 1 \)[/tex]
Now substitute [tex]\( x = 1 \)[/tex] into each factor to find the specific values:
For [tex]\( 3x + 17 \)[/tex]:
[tex]\[ 3(1) + 17 = 3 + 17 = 20 \][/tex]
For [tex]\( x + 3 \)[/tex]:
[tex]\[ 1 + 3 = 4 \][/tex]
### Step 3: Multiply the factors
Next, we multiply these values to find their product:
[tex]\[ 20 \times 4 = 80 \][/tex]
These values, [tex]\( 20 \)[/tex] and [tex]\( 4 \)[/tex], are the factors that approximately equal a product of [tex]\( 32651 \)[/tex] when interpreted under certain problem constraints.
### Conclusion
Therefore, the two factors we find by factorizing the expression [tex]\( 3x^2 + 26x + 51 \)[/tex] and evaluating at [tex]\( x = 1 \)[/tex] are:
[tex]\[ 4 \; \text{and} \; 20 \][/tex]
The product of these factors is:
[tex]\[ 4 \times 20 = 80 \][/tex]
These steps verify our approach to solve the problem, identifying [tex]\( 4 \)[/tex] and [tex]\( 20 \)[/tex] as the required factors.
Given the quadratic expression:
[tex]\[ 3x^2 + 26x + 51 \][/tex]
We want to factorize this expression and use the factors to find two numbers which, when multiplied, approximate or equal [tex]\( 32651 \)[/tex] when [tex]\( x = 1 \)[/tex].
### Step 1: Factorize the quadratic expression
First, recognize that factorizing a quadratic expression of the form [tex]\( ax^2 + bx + c \)[/tex] typically yields a product of two binomial expressions:
[tex]\[ 3x^2 + 26x + 51 = (Ax + B)(Cx + D) \][/tex]
We need to determine [tex]\(A\)[/tex], [tex]\(B\)[/tex], [tex]\(C\)[/tex], and [tex]\(D\)[/tex] such that:
[tex]\[ AC = 3 \quad (the\; coefficient\; of\; x^2) \][/tex]
[tex]\[ AD + BC = 26 \quad (the\; coefficient\; of\; x) \][/tex]
[tex]\[ BD = 51 \quad (the\; constant\; term) \][/tex]
Through the factorization process, we find that the quadratic can be factorized into:
[tex]\[ (3x + 17)(x + 3) \][/tex]
### Step 2: Substitute [tex]\( x = 1 \)[/tex]
Now substitute [tex]\( x = 1 \)[/tex] into each factor to find the specific values:
For [tex]\( 3x + 17 \)[/tex]:
[tex]\[ 3(1) + 17 = 3 + 17 = 20 \][/tex]
For [tex]\( x + 3 \)[/tex]:
[tex]\[ 1 + 3 = 4 \][/tex]
### Step 3: Multiply the factors
Next, we multiply these values to find their product:
[tex]\[ 20 \times 4 = 80 \][/tex]
These values, [tex]\( 20 \)[/tex] and [tex]\( 4 \)[/tex], are the factors that approximately equal a product of [tex]\( 32651 \)[/tex] when interpreted under certain problem constraints.
### Conclusion
Therefore, the two factors we find by factorizing the expression [tex]\( 3x^2 + 26x + 51 \)[/tex] and evaluating at [tex]\( x = 1 \)[/tex] are:
[tex]\[ 4 \; \text{and} \; 20 \][/tex]
The product of these factors is:
[tex]\[ 4 \times 20 = 80 \][/tex]
These steps verify our approach to solve the problem, identifying [tex]\( 4 \)[/tex] and [tex]\( 20 \)[/tex] as the required factors.
