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Factorise fully: Divide exercise between 2 days (1.) \( 3 a^{2}+10 a+3 \)

Ask by Barber Flynn. in South Africa
Feb 03,2025

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Tutor-Verified Answer

Answer

\( 3a^{2} + 10a + 3 = (3a + 1)(a + 3) \)

Solution

To factorise the quadratic expression \( 3a^{2} + 10a + 3 \) fully, follow these steps: ### Step 1: Identify Coefficients The given quadratic is in the form \( ax^{2} + bx + c \), where: - \( a = 3 \) - \( b = 10 \) - \( c = 3 \) ### Step 2: Find Two Numbers We need to find two numbers that multiply to \( a \times c = 3 \times 3 = 9 \) and add up to \( b = 10 \). The numbers **9** and **1** satisfy this: - \( 9 \times 1 = 9 \) - \( 9 + 1 = 10 \) ### Step 3: Split the Middle Term Rewrite the middle term using the numbers found: \[ 3a^{2} + 9a + 1a + 3 \] ### Step 4: Factor by Grouping Group the terms in pairs: \[ (3a^{2} + 9a) + (1a + 3) \] Factor out the greatest common factor from each group: \[ 3a(a + 3) + 1(a + 3) \] ### Step 5: Factor Out the Common Binomial Now, factor out the common binomial \( (a + 3) \): \[ (3a + 1)(a + 3) \] ### Final Answer \[ 3a^{2} + 10a + 3 = (3a + 1)(a + 3) \]

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To factorise the quadratic expression \( 3a^{2} + 10a + 3 \), we look for two numbers that multiply to \( 3 \times 3 = 9 \) (the product of the leading coefficient and the constant term) and add up to \( 10 \) (the coefficient of \( a \)). The numbers \( 1 \) and \( 9 \) fit this requirement! We can then split the middle term \( 10a \) into \( 1a + 9a \): \[ 3a^{2} + 1a + 9a + 3 \] Now, we can group the terms: \[ (3a^{2} + 1a) + (9a + 3) \] Factoring out the common terms gives: \[ a(3a + 1) + 3(3a + 1) \] Now notice \( (3a + 1) \) is common: \[ (3a + 1)(a + 3) \] So, \( 3a^{2} + 10a + 3 \) fully factors into \( (3a + 1)(a + 3) \)! Enjoy your math adventure!

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