Find the product of (3x – 1) and (10x2 – 19x + 6) and also verify

1.	Find the product of (3x – 1) and (10x2 – 19x + 6) and also verify the result for x = 2.
Answer» $The \: product \: of \: (3x-1) \:and \\(10x^{<klux>2</klux>}-19x+6)$ $= (3x-1)( 10x^{2}-19x+6)$ $=3x( 10x^{2}-19x+6) -1(10x^{2}-19x+6)$ $= 30x^{3} - 57x^{2} + 18x - 10x^{2} + 19x - 6$ $= 30x^{3} -(57+10)<klux>X</klux>^{2} + (18+19)x -6$ $= 30x^{3} -67x^{2} + 37x -6$ 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*/ $LHS \\= (3x-1)( 10x^{2}-19x+6)$ $= (3\times 2-1)[ 10\times 2^{2} - 19\times 2 + 6 ]\\=(6-1)(40-38+6)\\= 5\times 8\\= 40$ $RHS \\= 30x^{3} -67x^{2} + 37x -6$ $= 30\times 2^{3} - 67 \times 2^{2} + 37 \times 2 - 6 \\= 30 \times 8 - 67 \times 4 + 74 - 6\\= 240 - 268+68 \\= 40$ $LHS = RHS$ •••♪