Math, asked by anita223, 11 months ago

தொகு‌திமுறை வகு‌த்தலை‌ப் பய‌ன்படு‌த்‌தி ஈவு ம‌ற்று‌ம் ‌மீ‌தியை‌க் கா‌ண்க

i)(3x^3-2x^2+7x-5)÷(x+3) ii)8x^4-2x^2+6x+5)÷(4x+1)

Answers

Answered by steffiaspinno
0

தொகு‌திமுறை வகு‌த்தல்:

( i ) (3x^3 - 2x^2 + 7x - 5) ÷ ( x + 3)

தீர்வு:  

p(x) = (3x^3 - 2x^2 + 7 x - 5)

d (x) = x + 3, x = -3

Refer the diagram   (1)

ஈவு  : 3x^3 -11x + 40

மீதி : -125

(ii) 8x^4 - 2x^2 + 6 x + 5) ÷ (4x + 1)

தீர்வு:  

p(x) = (8 x^4 -2x^2 + 6x +5)

      =  8x^4 + 0x^3 - 2x^2 + 6x + 5

d( x ) = 4x + 1,

 x = \frac{-1}{4}

Refer the diagram   (2)

÷ 4

ஈவு  : 2x^3 - \frac{2}{4}x^2 - \frac{3}{8} + \frac{51}{52}

மீதி : \frac{109}{32}

Attachments:
Answered by Anonymous
0

\Large{\underline{\underline{\bf{Solution :}}}}

\rule{150}{2}

(i) d(x) = x + 3

x = -3

Let 3x³ - 2x² + 7x - 5 be equation (1).

Put Value of x in equation (1).

\sf{→3(-3)^3 - 2(-3)^2 + 7(-3) - 5} \\ \\ \sf{→3(-27) - 2(9) - 21 - 5} \\ \\ \sf{→-81 - 18 - 21 - 5} \\ \\ \sf{→-125}

\rule{200}{2}

(i) d(x) = 4x + 1

x = \frac{-1}{4}

Let 8x⁴ - 2x² + 6x + 5 be equation (2).

Put Value of x in equation (2).

\sf{→8(\frac{-1}{4})^4 + 2(\frac{-1}{4})^2 + \cancel{6}(\frac{-1}{\cancel{4}}) + 5} \\ \\ \sf{→\frac{\cancel{8}}{\cancel{256}} + \frac{\cancel{2}}{\cancel{16}} - \frac{3}{2} + 5} \\ \\ \sf{→\frac{1}{32} + \frac{1}{8} - \frac{3}{2} + 5} \\ \\ \sf{→\frac{1 + 4 - 48 + 160}{32}} \\ \\ \sf{→\frac{117}{32}}

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