GCSE OCR Maths More factorising - quadratics: (x + a)(x + 3)(2x + 1) = bx^3 +

(x + a)(x + 3)(2x + 1) = bx^3 + cx^2 + dx - 12 Find the value of a, b, c and d. - OCR - GCSE Maths - Question 17 - 2018 - Paper 1

Question 17

(x + a)(x + 3)(2x + 1) = bx^3 + cx^2 + dx - 12 Find the value of a, b, c and d.

Worked Solution & Example Answer:(x + a)(x + 3)(2x + 1) = bx^3 + cx^2 + dx - 12 Find the value of a, b, c and d. - OCR - GCSE Maths - Question 17 - 2018 - Paper 1

Step 1

96%

114 rated

Answer

First, expand the left-hand side of the equation:

Distribute (x + a)(x + 3):
- Using FOIL:
$(x + a)(x + 3) = x^2 + 3x + ax + 3a = x^2 + (3 + a)x + 3a$
Now, distribute this result with (2x + 1):

$(x^2 + (3 + a)x + 3a)(2x + 1)$
Expanding gives:
- First term: $2x^3 + (3 + a)2x^2 + 3a(2x)$
- Second term: $x^2 + (3 + a)x + 3a$

Thus, $= 2x^3 + (3 + 2a + 1)x^2 + (6a + 3)x + 3a$

Step 2

99%

104 rated

Answer

Writing the polynomial in standard form gives:

$2x^3 + (2a + 4)x^2 + (6a + 3)x + 3a - 12$

To identify b, c, and d, we compare coefficients:
b = 2, c = 2a + 4, d = 6a + 3 + 3a - 12 = 9a - 9$$

From our expression, by matching terms, we find that:

Step 3

96%

101 rated

Answer

Substituting the value of a back into c:

$c = 2(- rac{16}{9}) + 4 = - rac{32}{9} + rac{36}{9} = rac{4}{9}$

Step 4

98%

120 rated

Answer

Finally, knowing a = -rac{16}{9}:

Thus, we find:

b = 2, \ c = rac{4}{9}, \ d = -25$$