Question

Find the product of (3x – 1) and (10x2 – 19x + 6) and also verify the result for x = 2.​

Answer 1

$The \: product \: of \: (3x-1) \:and \\(10x^{2}-19x+6)$

$= (3x-1)( 10x^{2}-19x+6)$

$=3x( 10x^{2}-19x+6) -1(10x^{2}-19x+6)$

$= 30x^{3} - 57x^{2} + 18x - 10x^{2} + 19x - 6$

$= 30x^{3} -(57+10)x^{2} + (18+19)x -6$

$= 30x^{3} -67x^{2} + 37x -6$

Therefore.,

$\red{(3x-1)( 10x^{2}-19x+6)}$

$\green { = 30x^{3} -67x^{2} + 37x -6}$

Verification:

/* Substitute x = 2 in the above equation ,we get*/

$LHS \\= (3x-1)( 10x^{2}-19x+6)$

$= (3\times 2-1)[ 10\times 2^{2} - 19\times 2 + 6 ]\\=(6-1)(40-38+6)\\= 5\times 8\\= 40$

$RHS \\= 30x^{3} -67x^{2} + 37x -6$

$= 30\times 2^{3} - 67 \times 2^{2} + 37 \times 2 - 6 \\= 30 \times 8 - 67 \times 4 + 74 - 6\\= 240 - 268+68 \\= 40$

$LHS = RHS$

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Answer 2

Step-by-step explanation:

See the attached image above......

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