Steps | 6x2 + 13x + 5 = (2x + 1)(3x + 5) | ||
1. | Write the expression in equated form, with the factors in one side, preferable LHS, and the product on the other, preferably RHS. | In this case, (2x + 1)(3x + 5) = 6x2 + 13x + 5 |
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2. | Find the product of the (Sum of all the co-efficients) of the factors. | In this case, LHS = (2 + 1) × (3 + 5) = 3 × 8 = 24 |
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3. | Find the sum of all the co-efficients of the product expression. | In this case, RHS = 6 + 13 + 5 = 24 |
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4. | These two sums will always be the same. If, not the factors are incorrect and the expression cannot be equated. | In this case, since LHS = RHS, the expression in equated! |
The above expression is not equated because: Co-efficients of LHS = (2 + 3)(1 - 2) = -5 Co-efficients of RHS = 2 - 5 - 6 = -9 Thus, (2x + 3)(x - 2) ≠ 2x2 - 5x - 6 |
Co-efficients of LHS = (1 + 2)(1 + 3)(1 + 8) = 108 Co-efficients of RHS = 1 + 13 + 44 + 48 = 106 Thus, (x + 2)(x + 3)(x + 8) ≠ x3 + 13x2 + 44x + 48 |
Co-efficients of LHS = (1 + 1)(1 + 2)(1 + 3) = 24 Co-efficients of RHS = 1 + 6 + 11 + 6 = 24 Thus, (x + 1)(x + 2)(x + 3) = x3 + 6x2 + 11x + 6, is equated. |
For any Quadratic expression: (ax + b)(cx + d) = ac x2 + bc x + ad x + bd Considering only the co-efficients: ac + bc + ad + bd = a(c + d) + b(c + d) = (a + b)(c + d) Note, that Cubic and Bi-Quadratic expressions may be proven similarly. |
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