Question

factorise x³-23x²+142x-120

Answer 1

first take x=-1
so x+1=0
than put value 1 in x


than divide it by x-1



what answer came
divedend=divisor×quotient

Answer 2

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$\Huge{\red{\underline{\textsf{Answer}}}}$

Let p(x) = $\sf x^{3} - 23x^{2} + 142x - 120$

checking p(x) =

$\begin{tabular}{|c |c | c|}\cline{1-3}x & x^{3} - 23x^{2} + 142x - 120 </p><p>& whether 0 or not \\\cline{1-3}0 & -120 & No \\1 & 0 & yes \\\cline {1-3}\end{tabular}$

So at x = 1, p(x) = 0

Hence x - 1 is factor of p(x)

Now,

p(x) = (x - 1) g(x)

g(x) = $\large\sf \frac{p(x)}{(x - 1)}$

$\therefore$ g(x) is obtained after dividing p(x) by x - 1

$\boxed{\begin{array}{l | n | r}\sf x-1&\sf x^{3} - 23x^{2} + 142x - 120 &\sf x^2-22x+120\\ &\sf x^3-x^2\\ & ( - )\:( + )\\&\rule{100}{0.8}\\&\sf\qquad-22x^2+142x-120\\ &\sf\qquad-22x^2+22x\\ &\qquad( + )\:\:( - )\\&\quad\rule{100}{0.8}\\&\qquad\qquad\sf 120x-120\\ &\sf\qquad120x-120\\ &\qquad( - )\:\:( + )\\&\quad\rule{100}{0.8}\\&\qquad\qquad\sf 0\end{array}}$

So, g(x) = x² - 22x + 120

$\longrightarrow$ p(x) = (x - 1) g(x)

$\longrightarrow$ (x - 1) (x² - 22x + 120)

We factorize g(x) i.e x² - 22x + 120

$\leadsto$ $\sf x^2 - 22x + 120$ (by spliting middle term)

$\leadsto$ $\sf x^2 - 12x - 10x + 120$

$\leadsto$ $\sf x(x - 12) - 10(x - 12)$

$\leadsto$ $\sf (x - 12) (x - 10)$

So, p(x) = (x - 1)(x - 10)(x - 12)

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