Solve (x+?)(x+?) = x²+7x+10: Finding Missing Terms in Quadratic Factors

Q: How do I know which numbers to look for?

Look at the constant term (10) for the product and the middle coefficient (7) for the sum. You need two numbers where: number₁ × number₂ = 10 and number₁ + number₂ = 7.

Q: Why can't I use (x+1)(x+10)?

Let's check: $ (x+1)(x+10) = x^2 + 11x + 10 $. The middle term is 11x, not 7x! Always verify your factors give the correct middle term.

Q: Do I always write the larger number first?

No! The order doesn't matter for the final answer. $ (x+2)(x+5) $ and $ (x+5)(x+2) $ are both correct since multiplication is commutative.

Quadratic Factoring with Missing Terms

$(x+?)(x+?)=x^2+7x+10$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Complete the appropriate numbers

00:09 Find the trinomial coefficients

00:16 We want to find 2 numbers whose sum equals B

00:21 and their product equals C

00:29 These are the appropriate numbers, let's substitute in the trinomial

00:35 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$(x+?)(x+?)=x^2+7x+10$

Step-by-step solution

To solve this problem, we'll follow these steps:

Step 1: Identify the pattern of factoring quadratics.
Step 2: Use the sum and product relation to find the numbers.
Step 3: Confirm the binomial expressions.

Let's work through each step:

Step 1: Our task is to factor the expression $x^2 + 7x + 10$ . We need two numbers whose product is $10$ (the constant term) and whose sum is $7$ (the coefficient of $x$ ).

Step 2: We list pairs of numbers that multiply to $10$ :
- $1 \times 10$
- $2 \times 5$

Among these, only the pair $2$ and $5$ add up to $7$ . Therefore, the expressions needed are $(x+2)$ and $(x+5)$ .

Step 3: We substitute these values into the binomials: $(x+5)(x+2)$ . Expanding this verifies:
$(x+5)(x+2) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10$ .

Therefore, the factorization is correct, and the solution reached is $\left(x+5\right)\left(x+2\right)$ .

Therefore, the solution to the problem is $\left(x+5\right)\left(x+2\right)$ .

Final Answer

$\left(x+5\right)\left(x+2\right)$

Key Points to Remember

Essential concepts to master this topic

Sum-Product Rule: Find two numbers that add to 7 and multiply to 10
Factor Pairs: Check 1×10 and 2×5, only 2+5=7 works
Verification: Expand (x+5)(x+2) = x²+7x+10 to confirm ✓

Common Mistakes

Avoid these frequent errors

Using factor pairs that multiply correctly but don't add correctly
Don't just find numbers that multiply to 10 like 1×10 without checking if they add to 7! This gives wrong factors like (x+1)(x+10) = x²+11x+10 instead of x²+7x+10. Always verify both the sum AND product conditions.

Practice Quiz

Test your knowledge with interactive questions

$(3+20)\times(12+4)=$

FAQ

Everything you need to know about this question

How do I know which numbers to look for?

Look at the constant term (10) for the product and the middle coefficient (7) for the sum. You need two numbers where: number₁ × number₂ = 10 and number₁ + number₂ = 7.

What if there are negative signs in the quadratic?

The signs matter! If the constant is positive like +10, both numbers have the same sign. If the middle term is positive like +7x, both numbers are positive.

Why can't I use (x+1)(x+10)?

Let's check: $(x+1)(x+10) = x^2 + 11x + 10$ . The middle term is 11x, not 7x! Always verify your factors give the correct middle term.

Is there a faster way to find the numbers?

List all factor pairs of the constant term first: for 10, that's 1×10 and 2×5. Then quickly check which pair adds to the middle coefficient.

What if no factor pairs work?

If no integer pairs multiply to give the constant and add to give the middle coefficient, then the quadratic doesn't factor nicely with integers. You'd need other methods like the quadratic formula.

Do I always write the larger number first?

No! The order doesn't matter for the final answer. $(x+2)(x+5)$ and $(x+5)(x+2)$ are both correct since multiplication is commutative.

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