Applying C1→ C1 + C2 + C3
Taking (5x + 4) common from C1
Apply R2→ R2 – R1 and R3→ R3 – R1
Expanding along C1, we get
= (5x + 4)[1(4 – x)(4 – x) – 0(0)]
= (5x + 4)(4 – x)2
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