Factorize: x^2 - 7x + 12. | Mathematics MCQ | Examoogle

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Factorize: x² - 7x + 12.

(x-3)(x-4)

(x+3)(x+4)

(x-2)(x-6)

(x+2)(x+6)

Correct Answer

📐 MA • Math Direct FormulaMathematicsAlgebra

Option A

(x-3)(x-4)

Quick Summary:

Look for two numbers that multiply to give the constant term ( $+12$ ) and add up to give the coefficient of the middle term ( $-7$ ). These numbers are $-3$ and $-4$ .

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📐MAMath SolutionDirect Formula

📋 Given

Quadratic expression: x² - 7x + 12

🔢 Formula Used

x2−(a+b)x+ab=(x−a)(x−b)x^2 - (a + b)x + ab = (x - a)(x - b)x2−(a+b)x+ab=(x−a)(x−b)

⚡ Exam Hall Shortcut / Speed Trick

Look for two numbers that multiply to give the constant term ( $+12$ ) and add up to give the coefficient of the middle term ( $-7$ ). These numbers are $-3$ and $-4$ .

⚠️ Common Student Trap / Pitfall

Choosing numbers with incorrect signs, such as $+3$ and $+4$ , which would give a middle term of $+7x$ instead of $-7x$ .

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Identify coefficients

Compare the given quadratic expression $x^2 - 7x + 12$ with the standard form $ax^2 + bx + c$ , where $a = 1$ , $b = -7$ , and $c = 12$ .

ax2+bx+cax^2 + bx + cax2+bx+c

Find factors of the constant term

We need to find two numbers whose product is $12$ and whose sum is $-7$ .

p×q=12,p+q=−7p \times q = 12, \quad p + q = -7p×q=12,p+q=−7

Determine the numbers

The factors of $12$ are $(1, 12), (2, 6), (3, 4)$ . Taking negative signs, $-3$ and $-4$ multiply to $+12$ and add up to $-7$ .

−3×(−4)=12,(−3)+(−4)=−7-3 \times (-4) = 12, \quad (-3) + (-4) = -7−3×(−4)=12,(−3)+(−4)=−7

Write the factored form

Substitute the numbers into the factored form expression to obtain $(x - 3)(x - 4)$ .

(x−3)(x−4)(x - 3)(x - 4)(x−3)(x−4)

✅

A is correct because factorizing $x^2 - 7x + 12$ yields the factors $(x-3)(x-4)$ .

Core Concepts Used

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Quadratic Equations Factorization by Splitting the Middle Term Algebraic Identities

💡 EXAM TIP

Mastering factorization is essential for quickly solving quadratic equations and finding roots in competitive exams without using the quadratic formula.

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