To factor the quadratic expression \(3z^2 - 4z - 7\), we can use the method of factoring by grouping or the quadratic formula. However, in this case, it might not factor nicely with integers.
We can first check if the quadratic can be factored into the form \((az + b)(cz + d)\).
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The product of \(ac\) (where \(a\) is the coefficient of \(z^2\) and \(c\) is the constant term) is \(3 \times -7 = -21\).
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We need to find two numbers that multiply to give \(-21\) and add up to the coefficient of \(z\) which is \(-4\). Those two numbers are \(3\) and \(-7\).
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We can rewrite the middle term \(-4z\) using those numbers: \[ 3z^2 + 3z - 7z - 7 \]
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Now we can factor by grouping: \[ (3z^2 + 3z) + (-7z - 7) \] Factor out the common terms in each group: \[ 3z(z + 1) - 7(z + 1) \]
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Now we can factor out the common binomial \((z + 1)\): \[ (3z - 7)(z + 1) \]
So, the expression \(3z^2 - 4z - 7\) factors into: \[ (3z - 7)(z + 1) \]
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