Factoring the Quadratic Function: Find the Product of x²+11x+28

Quadratic Factoring with Positive Coefficients

Find the representation of the product of the following function

f(x)=x2+11x+28 f(x)=x^2+11x+28

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Convert trinom to multiplication
00:04 Identify the trinom coefficients
00:08 We want to find 2 numbers whose sum equals coefficient B
00:12 and their product equals coefficient C
00:20 These are the matching numbers, we'll substitute in multiplication
00:27 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the representation of the product of the following function

f(x)=x2+11x+28 f(x)=x^2+11x+28

2

Step-by-step solution

To solve this problem, we need to find a product representation of the quadratic function f(x)=x2+11x+28 f(x) = x^2 + 11x + 28 .

Let's go through the problem-solving process step-by-step:

1. We need to express the given quadratic function f(x)=x2+11x+28 f(x) = x^2 + 11x + 28 as a product of two binomials. 2. Key information includes the coefficients of the terms in the quadratic: 1 for x2 x^2 , 11 for x x , and 28 for the constant term. 3. The approach is to factor the quadratic expression. 4. Key formula: For a quadratic expression ax2+bx+c ax^2 + bx + c , the product of factors (x+p)(x+q) (x+p)(x+q) requires that p+q=b/a p+q = b/a and pq=c/a pq = c/a . 5. Factoring by finding two numbers whose sum is 11 and product is 28. 6. Steps: Find two numbers that satisfy the conditions p+q=11 p+q = 11 and pq=28 pq = 28 . 7. Assumptions: The quadratic can be factored into real numbers. 8. Consider pairs of factors of 28: (1,28),(2,14),(4,7),(−1,−28),(−2,−14),(−4,−7) (1, 28), (2, 14), (4, 7), (-1, -28), (-2, -14), (-4, -7) . 9. Edge cases: Check that potential factors satisfy both conditions. 10. We consider only the choices since the problem is a multiple-choice. 11. For multiple-choice questions, find the correct pair of numbers by testing the conditions. 12. Common mistakes might include not correctly identifying factor pairs or choosing incorrect signs. 13. Changes in coefficients would alter the factor pairs needed to produce the correct sum and product.

Now let's address the problem:

To find the correct binomial factors of f(x)=x2+11x+28 f(x) = x^2 + 11x + 28 , we are searching for two numbers that multiply to 28 and add to 11. Examining possible pairs, 4 4 and 7 7 meet these criteria: 4×7=28 4 \times 7 = 28 and 4+7=11 4 + 7 = 11 .

Thus, the quadratic expression can be rewritten as:

f(x)=(x+4)(x+7) f(x) = (x + 4)(x + 7)

This checks our desired conditions. Multiplication confirms: (x+4)(x+7)=x2+7x+4x+28=x2+11x+28(x+4)(x+7) = x^2 + 7x + 4x + 28 = x^2 + 11x + 28, which matches the original function.

Therefore, the product representation of f(x)=x2+11x+28 f(x) = x^2 + 11x + 28 is (x+7)(x+4) (x+7)(x+4) .

3

Final Answer

(x+7)(x+4) (x+7)(x+4)

Key Points to Remember

Essential concepts to master this topic
  • Factor Rule: Find two numbers that multiply to 28 and add to 11
  • Technique: Test factor pairs: 4×7=28 and 4+7=11 works perfectly
  • Check: Expand (x+4)(x+7) = x²+11x+28 matches original ✓

Common Mistakes

Avoid these frequent errors
  • Getting the signs wrong in factors
    Don't write (x-4)(x-7) when all coefficients are positive = negative middle term! When the constant term is positive and the middle coefficient is positive, both factors must have positive signs. Always check: if c>0 and b>0, then both factor signs are positive.

Practice Quiz

Test your knowledge with interactive questions

Create an algebraic expression based on the following parameters:

\( a=3,b=0,c=-3 \)

FAQ

Everything you need to know about this question

How do I know which factor pairs of 28 to try?

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List all factor pairs systematically: (1,28), (2,14), (4,7). Then check which pair adds up to the middle coefficient (11). Only 4+7=11, so those are your numbers!

Why can't the answer be (x-7)(x-4)?

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Because expanding (x−7)(x−4) (x-7)(x-4) gives x2−11x+28 x^2 - 11x + 28 , not x2+11x+28 x^2 + 11x + 28 . The signs matter - negative factors give a negative middle term!

What if I can't find two numbers that work?

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First, double-check your factor pairs and arithmetic. If you've tried all possibilities and nothing works, the quadratic might not factor nicely with integers - but for this problem, 4 and 7 definitely work!

Does the order of factors matter?

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No! (x+4)(x+7) (x+4)(x+7) is the same as (x+7)(x+4) (x+7)(x+4) because multiplication is commutative. Both are correct answers.

How can I check my factoring without fully expanding?

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Quick check: multiply the constant terms (4×7=28 ✓) and add the outer coefficients (4+7=11 ✓). If both match your original quadratic, you're right!

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