Fully factorise c^2 + 3c - 28
Understand the Problem
The question asks to fully factor the quadratic expression c^2 + 3c - 28. This involves finding two binomials that, when multiplied, give the original quadratic expression.
Answer
$(c + 7)(c - 4)$
Answer for screen readers
$(c + 7)(c - 4)$
Steps to Solve
- Identify the coefficients
The quadratic expression is in the form $ac^2 + bc + c$, where $a = 1$, $b = 3$, and $c = -28$.
- Find two numbers that multiply to $ac$ and add up to $b$
We need two numbers that multiply to $1 \times -28 = -28$ and add up to $3$. The numbers are $7$ and $-4$ because $7 \times -4 = -28$ and $7 + (-4) = 3$.
- Rewrite the middle term using the found numbers
Rewrite the middle term $3c$ as $7c - 4c$. The expression becomes $c^2 + 7c - 4c - 28$.
- Factor by grouping
Group the first two terms and the last two terms: $(c^2 + 7c) + (-4c - 28)$. Factor out the greatest common factor (GCF) from each group: $c(c + 7) - 4(c + 7)$.
- Factor out the common binomial
Notice that $(c + 7)$ is a common factor. Factor it out: $(c + 7)(c - 4)$.
$(c + 7)(c - 4)$
More Information
The factored form of the quadratic $c^2 + 3c - 28$ is $(c + 7)(c - 4)$, meaning if you multiply $(c + 7)$ by $(c - 4)$, you will retrieve the original quadratic expression.
Tips
A common mistake is getting the signs wrong when finding the two numbers that multiply to $ac$ and add up to $b$. For example, students might incorrectly choose $-7$ and $4$ instead of $7$ and $-4$.
Another mistake is not factoring out the negative sign correctly when grouping, which can lead to incorrect binomial factors and a wrong final answer.
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