Answer:
The prime factors of q = 330 are 2, 3, 5 and 11.
HCF of [tex](6x^2+ x -15)[/tex] and [tex](8x^2-32x + 15)[/tex] is 1.
Step-by-step explanation:
First of all, let us convert the given decimal expansion in rational form.
Let Given decimal expansion:
[tex]x =0.27575.....[/tex]
Multiply by 100:
[tex]100x = 27.5757.......[/tex]
subtracting x from 100x :
[tex]99x = 27.5757.... - 0.2757....\\\Rightarrow 99x = 27.3\\\Rightarrow 33x = 9.1\\\Rightarrow x = \dfrac{9.1}{33}\\\Rightarrow x = \dfrac{91}{330}[/tex]
So,
[tex]\dfrac{p}{q} = \dfrac{91}{330}[/tex]
Let us make prime factors of q:
[tex]q = 2 \times 3 \times 5\times 11[/tex]
The prime factors of q = 330 are 2, 3, 5 and 11.
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Finding HCF of [tex](6x^2+ x -15)[/tex] and [tex](8x^2-32x + 15)[/tex]
Method to find HCF:
First of all, let us factorize them and the common part between the factors of the two will be the HCF.
[tex](6x^2+ x -15)\\\Rightarrow 6x^2+ 10x -9x-15\\\Rightarrow 2x(3x+ 5) -3(3x+5)\\\Rightarrow (2x -3)(3x+5)[/tex]
The factors are (2x -3) and (3x+5)
The quadratic equation [tex](8x^2-32x + 15)[/tex] can not be factorized as we have factorized [tex](6x^2+ x -15)[/tex].
[tex]\therefore[/tex] HCF of the two is 1.
