Question

Without actual division prove that x4 -5x3 +8x2 -10x+12 is divisible by x-5x +6.​

Answer 1

Step-by-step explanation:

x-5+6+7,y-8 is your correct Answer

hope it's help you

Answer 2

Answer:

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$\underline{\underline{\sf{\green{To\:Prove\::-}}}}$

x⁴ - 5x³ + 8x² - 10x + 12 is divisible by x² - 5x + 6.

$\underline{\underline{\sf{\green{Proof\::-}}}}$

$\small\rightarrow\:\sf p(x) = x^4 - 5x^3 + 8x^2 - 10x + 12 \:\bigg\lgroup eq^{n}\: (1)\bigg\rgroup$

$\small\rightarrow\:\sf x^2 + 5x + 6 = 0\:\qquad\quad\qquad\:\:\qquad\bigg\lgroup eq^{n}\: (2)\bigg\rgroup$

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★ Splitting the middle term of [eqⁿ (2)] :-

$\rightarrow\quad\sf x^2 - 2x - 3x + 6 = 0$

$\rightarrow\quad\sf x(x - 2) - 3(x - 2) = 0$

$\rightarrow\quad\sf (x - 2)(x - 3) = 0$

$\rightarrow\quad\sf x - 2 = 0\:,\:x - 3 = 0$

$\rightarrow\quad{\bf{\pink{x = 2\:,\:x = 3}}}$

★ Putting x = 2, in [eqⁿ (1)] :-

$\rightarrow\quad\sf p(x) = (2)^4 - 5(2)^3 + 8(2)^2 - 10(2) + 12$

$\rightarrow\quad\sf p(x) = 16 - 5(8) + 8(4) - 20 + 12$

$\rightarrow\quad\sf p(x) = 16 - 40 + 32 - 20 + 12$

$\rightarrow\quad\sf p(x) = 32 + 12 + 16 - 20 - 40$

$\rightarrow\quad\sf p(x) = 60 - 60$

$\rightarrow\quad\sf p(x) = \cancel{60} - \cancel{60}$

$\rightarrow\quad{\bf{\red{p(x) = 0}}}$

★ Putting x = 3, in [eqⁿ (1)] :-

$\rightarrow\quad\sf p(x) = (3)^4 - 5(3)^3 + 8(3)^2 - 10(3) + 12$

$\rightarrow\quad\sf p(x) = 81 - 5(27) + 8(9) - 30 + 12$

$\rightarrow\quad\sf p(x) = 81 - 135 + 72 - 30 + 12$

$\rightarrow\quad\sf p(x) = 72 + 12 + 81 - 135 - 130$

$\rightarrow\quad\sf p(x) = 165 - 165$

$\rightarrow\quad\sf p(x) = \cancel{165} - \cancel{165}$

$\rightarrow\quad{\bf{\purple{p(x) = 0}}}$

$\qquad\qquad\quad\underline{\underline{\sf{\green{Hence,\:Proved!}}}}$

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Both time, putting x = 2 and x = 3, in [eqⁿ (1)] we get p(x) = 0.

Therefore, x⁴ - 5x³ + 8x² - 10x + 12 is divisible by x² - 5x + 6.

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