Question

Solve for x.
x³ - 9x² + 26x - 24 = 0

Kindly don't comment irrelevant.

Thank you :)​

Answer 1

p(x) = x³- 9x² +26x - 24

1.Let x = 2

(2)³ - 9(2)² + 26(2) - 24

8 - 36 + 52 - 24 = 0

This means x = 2 is a solution of p(x) .

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2.Now let x = 3

(3)³ - 9(3)² + 26(3) - 24

9 - 81 + 78 - 24 = 0

This means x = 3 is also a solution of p(x)

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3.Let x = 4

(4)³ - 9(4)² + 26(4) - 24

64 - 144 + 104 - 24 = 0

This means x = 4 is also a solution of p(x)

From the above calculations we got ;

• x = 2

• x = 3

• x = 4.

Answer 2

$\sf{x^3-9x^2+26x-24=0}$

Use the Rational Root Theorem (Rational Root Theorem : For a polynomial equation with integer coefficients $\sf{a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}}$ . If $\sf{a_0}$ and $\sf{a_n}$ are integers. Then if there is a rational solution it could be found by checking all the numbers produced for ± Dividers of $\sf{a_0}$ ÷ Dividers of $\sf{a_n}$)

• Here, $\sf{a_0}$ = 24 and $\sf{a_n} = 1$

• The Dividers of $\sf{a_0: 1,\:2,\:3,\:4,\:6,\:8,\:12,\:24}$

• The Dividers of $\sf{a_n}$ = 1

Therefore Check the following rational numbers

$\sf{\displaystyle\pm \frac{1,\:2,\:3,\:4,\:6,\:8,\:12,\:24}{1}}$

• 2/1 is a root of the expression, so factor out x - 2

$\to\sf{\left(x-2\right)\dfrac{x^3-9x^2+26x-24}{x-2}}$

$\to\sf{\displaystyle(x-2)\left(x^2+\frac{-7x^2+26x-24}{x-2}\right)}$

$\to\sf{\displaystyle(x-2)\left(x^2-7x+\frac{12x-24}{x-2}\right)}$

$\to\sf{(x-2)(x^2-7x+12)}$

Let's Factor x² - 7x + 12 and continue solving

• We can write x² - 7x + 12 = x² - 3x - 4x + 12

$\to\sf{(x-2)(x^2-3x-4x+12)}$

$\to\sf{(x-2)[x(x-3)-4(x-3)]}$

• Factor out common term : x - 3

$\to\sf{(x-2)\left(x-3\right)\left(x-4\right)}$

Use Zero Factor Principle

• The Principle of Zero Products states that if the product of two numbers is 0, If ab = 0 then a = 0 or b = 0

$\boxed{\sf{x-2=0\quad \mathrm{or}\quad \:x-3=0\quad \mathrm{or}\quad \:x-4=0}}$

• Solve this three Equations one by one Now

$\bullet\sf{\:\:x-2=0\quad \implies x-2+2=0+2\quad \implies \underline{\underline{x=2}}}$

$\bullet\sf{\:\:x-3=0\quad \implies x-3+3=0+3\quad \implies \underline{\underline{x=3}}}$

$\bullet\sf{\:\:x-4=0\quad \implies x-4+4=0+4\quad \implies \underline{\underline{x=4}}}$

The Final Solutions are,

• $\sf{x = 2,\:x = 3, \:x = 4}$