Question

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

quality answer needed no spamming​

Answer 1

EXPLANATION.

Without actual division,

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

As we know that,

⇒ x² - 5x + 6 = 0.

Factorizes the equation into middle term splits, we get.

⇒ x² - 3x - 2x + 6 = 0.

⇒ x(x - 3) - 2(x - 3) = 0.

⇒ (x - 2)(x - 3) = 0.

⇒ x = 2 and x = 3.

Put the value of x = 2 in the equation, we get.

⇒ p(x) = x⁴ - 5x³ + 8x² - 10x + 12.

⇒ p(2) = (2)⁴ - 5(2)³ + 8(2)² - 10(2) + 12.

⇒ p(2) = 16 - 40 + 32 - 20 + 12.

⇒ p(2) = 16 + 32 + 12 - 40 - 20.

⇒ p(2) = 60 - 60.

⇒ p(2) = 0.

Put the value of x = 3 in the equation, we get.

⇒ p(x) = x⁴ - 5x³ + 8x² - 10x + 12.

⇒ p(3) = (3)⁴ - 5(3)³ + 8(3)² - 10(3) + 12.

⇒ p(3) = 81 - 135 + 72 - 30 + 12.

⇒ p(3) = 81 + 72 + 12 - 135 - 30.

⇒ p(3) = 165 - 165.

⇒ p(3) = 0.

Hence Proved.

Answer 2

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

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

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

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

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

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

★ 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)] :-

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

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

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

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

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

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

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

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

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

• 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.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬