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Question 1. Asked on :12 January 2019:10:49:51 AM

#### Factorise:x3-23x2+142x-120

-Added by Akki chauhan Mathematics » Polynomials

Himanshi Verma

x3 -23x2+142x-120

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Himanshi Verma On 13 January 2019:03:45:14 PM(220Average Rating Based on rating)

Akki chauhan

x-23x2+142x-120

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Akki chauhan On 14 January 2019:10:16:09 AM(341Average Rating Based on rating)

khushi chauhan

x-23x2+142x-120

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by khushi chauhan On 14 January 2019:06:03:32 PM(438Average Rating Based on rating)

Akki chauhan

x-23x2+142x-120

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Akki chauhan On 15 January 2019:09:04:11 AM(480Average Rating Based on rating)

Himanshi Verma

x-23x2+142x-120

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Himanshi Verma On 16 January 2019:05:57:48 PM(723Average Rating Based on rating)

Master Purushottam

x-23x2+142x-120

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Master Purushottam On 22 August 2019:09:12:12 PM(4314Average Rating Based on rating)

Shivang Gupta

x-23x2+142x-120

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Shivang Gupta On 25 August 2019:05:08:24 PM(6424Average Rating Based on rating)

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