To solve the quadratic equation \( x^2 - 13x + 42 = 0 \) by grouping, we need to split the middle term, \(-13x\), into two terms that add up to \(-13\) and multiply to \(42\) (the constant term).
We are looking for two numbers \(a\) and \(b\) such that:
- \(a + b = -13\)
- \(a \cdot b = 42\)
Let's consider the factors of \(42\):
- \(1 \times 42\)
- \(2 \times 21\)
- \(3 \times 14\)
- \(6 \times 7\)
Among these pairs, we need to find a pair that sums to \(-13\). By observing that both numbers need to be negative (since their product is positive and their sum is negative), we can check:
- \(-6\) and \(-7\) fit:
- \(-6 + (-7) = -13\)
- \(-6 \cdot -7 = 42\)
Thus, the numbers needed to split the middle term are \(-6\) and \(-7\).
So, we can rewrite the equation as: \[ x^2 - 6x - 7x + 42 = 0 \]
Then, we can group and factor accordingly.
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