GCSE OCR Maths Factorising: (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

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Answer

To start simplifying the left side of the equation, we will expand the expression step by step.

First, we'll multiply the first two brackets:

egin{align*} (x + a)(x + 3) &= x^2 + 3x + ax + 3a
&= x^2 + (3 + a)x + 3a

ewline ext{Now, we'll multiply this result by the last bracket, }(2x + 1):

ewline (x^2 + (3 + a)x + 3a)(2x + 1)

ewline = 2x^3 + x^2(3 + a) + 6ax + (3a)(2x + 1)

ewline = 2x^3 + (3 + a)x^2 + 6ax + 3a(2x) + 3a

ewline = 2x^3 + (3 + a + 6a)x^2 + 3a(2x + 1)

ewline = 2x^3 + (3 + 7a)x^2 + 6a + 3a.

ewline ext{This results in: } 2x^3 + (3 + 7a)x^2 + 3a

We now equate the coefficients from both sides of the equation with } bx^3 + cx^2 + dx - 12.

Step 2

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Answer

By comparing coefficients:

For the coefficient of $x^3$ : ( b = 2 )
For the coefficient of $x^2$ : ( c = 3 + 7a )
For the coefficient of $x$ : ( d = 6a + 3a ), which simplifies to ( d = 9a )
Lastly, to find the constant term, we set the constant on the right side equal to -12: ( 3a = -12 ) ( a = -4 )

Now substituting ( a = -4 ) back into the equations for ( b ), ( c ), and ( d ):

Thus, the final values are: