Algebra

Quadratic Equations

Easy

The factors of polynomial $x^{3}-6 x^{2}+11 x-6$ are :

$(x+1)(x+2)(x+3)$

$(x-1)(x+2)(x+3)$

$(x-1)(x-2)(x-3)$

$(x-1)(x-2)(x+3)$

✅ Correct Option: 3

Factors mean that each of these factors divide the polynomial perfectly. So, if we looked at Option 1, then:

$x+1=x+2=x+3=0,$ which would mean $x=-1,-2$ or $-3$ . These values, when put in the polynomial, should be equating 0. So, go by the options doing this. Look at Option 3.

Then, $x=1,2,3.$

Putting $x=1$

$(1)^3-6(1)^2+11(1)-6=1-6+11-6=0$

Putting $x=2$

$(2)^3-6(2)^2+11(2)-6=8-24+22-6=0$

Putting $x=3$

$(3)^3-6(3)^2+11(3)-6=27-54+33-6=0$

Hence, Option 3 is the answer.

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Solution

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JIPMAT Past Year Questions (Topic-Wise)

The factors of polynomial $x^{3}-6 x^{2}+11 x-6$ are :