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18 (a) Show that $(2x + 1)(x + 3)(3x + 7)$ can be written in the form $ax^3 + bx^2 + cx + d$ where $a, b, c$ and $d$ are integers - Edexcel - GCSE Maths - Question 19 - 2019 - Paper 3

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18-(a)-Show-that-$(2x-+-1)(x-+-3)(3x-+-7)$-can-be-written-in-the-form-$ax^3-+-bx^2-+-cx-+-d$-where-$a,-b,-c$-and-$d$-are-integers-Edexcel-GCSE Maths-Question 19-2019-Paper 3.png

18 (a) Show that $(2x + 1)(x + 3)(3x + 7)$ can be written in the form $ax^3 + bx^2 + cx + d$ where $a, b, c$ and $d$ are integers. (b) Solve $(1 - x)^2 < \frac{9}{2... show full transcript

Worked Solution & Example Answer:18 (a) Show that $(2x + 1)(x + 3)(3x + 7)$ can be written in the form $ax^3 + bx^2 + cx + d$ where $a, b, c$ and $d$ are integers - Edexcel - GCSE Maths - Question 19 - 2019 - Paper 3

Step 1

Show that $(2x + 1)(x + 3)(3x + 7)$ can be written in the form $ax^3 + bx^2 + cx + d$ where $a, b, c$ and $d$ are integers.

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Answer

To show that (2x+1)(x+3)(3x+7)(2x + 1)(x + 3)(3x + 7) can be expressed in the desired form:

  1. Expand the Expression Step-by-Step:

    First, expand (x+3)(3x+7)(x + 3)(3x + 7):

    (x+3)(3x+7)=3x2+7x+9x+21=3x2+16x+21(x + 3)(3x + 7) = 3x^2 + 7x + 9x + 21 = 3x^2 + 16x + 21

    Next, multiply this result by (2x+1)(2x + 1):

    (2x+1)(3x2+16x+21)=2x(3x2+16x+21)+1(3x2+16x+21)(2x + 1)(3x^2 + 16x + 21) = 2x(3x^2 + 16x + 21) + 1(3x^2 + 16x + 21)

    Expanding:

    =6x3+32x2+42x+3x2+16x+21= 6x^3 + 32x^2 + 42x + 3x^2 + 16x + 21

    Combining like terms gives:

    6x3+(32+3)x2+(42+16)x+21=6x3+35x2+58x+216x^3 + (32 + 3)x^2 + (42 + 16)x + 21 = 6x^3 + 35x^2 + 58x + 21

    Hence, we can write:

    a=6,b=35,c=58,d=21a = 6, b = 35, c = 58, d = 21

    All coefficients are integers, thus satisfying the requirement.

Step 2

Solve $(1 - x)^2 < \frac{9}{25}$

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Answer

To solve the inequality:

  1. Take the Square Root of Both Sides:

    1x<35|1 - x| < \frac{3}{5}

  2. Set Up Two Inequalities:

    This gives two cases:

    • Case 1:
      1x<351 - x < \frac{3}{5}
      Rearranging this gives:
      x>135=25x > 1 - \frac{3}{5} = \frac{2}{5}

    • Case 2:
      (1x)<35-(1 - x) < \frac{3}{5}
      This simplifies to:
      x1<35x - 1 < \frac{3}{5}
      And further rearranging gives:
      x<1+35=85x < 1 + \frac{3}{5} = \frac{8}{5}

  3. Combine Both Inequalities:

    Hence, the solution for xx is:

    25<x<85\frac{2}{5} < x < \frac{8}{5}

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