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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 20 - 2019 - Paper 3

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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 20-2019-Paper 3.png

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

Worked Solution & Example Answer: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 20 - 2019 - Paper 3

Step 1

Show that $(2x + 1)(x + 3)(3x + 7)$ can be written in the form $ax^3 + bx^2 + cx + d$

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Answer

To express the product (2x+1)(x+3)(3x+7)(2x + 1)(x + 3)(3x + 7) in the form ax3+bx2+cx+dax^3 + bx^2 + cx + d, we will perform the multiplication systematically.

  1. Multiply the first two factors:

    (2x+1)(x+3)=2x2+6x+x+3=2x2+7x+3(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3

  2. Now multiply this result by the third factor:

    (2x2+7x+3)(3x+7)(2x^2 + 7x + 3)(3x + 7)

  3. Use the distributive property (FOIL):

    • 2x2imes3x=6x32x^2 imes 3x = 6x^3
    • 2x2imes7=14x22x^2 imes 7 = 14x^2
    • 7ximes3x=21x27x imes 3x = 21x^2
    • 7ximes7=49x7x imes 7 = 49x
    • 3imes3x=9x3 imes 3x = 9x
    • 3imes7=213 imes 7 = 21

Combining like terms gives:

6x3+(14x2+21x2)+(49x+9x)+216x^3 + (14x^2 + 21x^2) + (49x + 9x) + 21

Simplifying further leads to: 6x3+35x2+58x+216x^3 + 35x^2 + 58x + 21.

Thus, comparing with the form ax3+bx2+cx+dax^3 + bx^2 + cx + d, we have:

  • a=6a = 6
  • b=35b = 35
  • c=58c = 58
  • d=21d = 21.

Step 2

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

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Answer

To solve the inequality (1x)2<925(1 - x)^2 < \frac{9}{25}, we start by taking the square root of both sides:

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

This leads us to two inequalities:

  1. 1x<351 - x < \frac{3}{5}
  2. 1x>351 - x > -\frac{3}{5}

For the first inequality:

1x<351 - x < \frac{3}{5}

Subtracting 1 from both sides:

x<351-x < \frac{3}{5} - 1

x<3555-x < \frac{3}{5} - \frac{5}{5}

x<25-x < -\frac{2}{5}

Multiplying by -1 (reversing the inequality):

x>25x > \frac{2}{5}

For the second inequality:

1x>351 - x > -\frac{3}{5}

Subtracting 1 gives:

x>351-x > -\frac{3}{5} - 1

x>3555-x > -\frac{3}{5} - \frac{5}{5}

x>85-x > -\frac{8}{5}

Multiplying by -1 (reversing the inequality):

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

Combining both parts, we find: 25<x<85\frac{2}{5} < x < \frac{8}{5}.

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