Math, asked by shyamalihazra16123, 8 months ago

Show that (x-1) is a factor of x^3-7x^2+15x -9
Hence
factorise

Answers

Answered by jtg07

Step-by-step explanation:

if (x-1) is a factor of the given polynomial...then x=1 would be a zero of the given polynomial so let's test it ...

put x=1 in equation...

$( {1}^{3} - 7( {1}^{2} ) + 15(1) - 9)$

$1 - 7 + 15 - 9$

$16 - 16$

$0$

thus x-1 is a factor of the given polynomial....

to find rest zeros we have to divide the factor from the given polynomial....the division has been shown in the file above please refer to it....

so we get the polynomial x²-6x+9 after division....after factorising it....

${x}^{2} - 6x + 9$

${x}^{2} - 3x - 3x + 9$

$x(x - 3) - 3(x - 3)$

$(x - 3)(x - 3)$

so the factors are x=3,3,1

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Answered by Anonymous

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Show that (x-1) is a factor of x^3-7x^2+15x -9

So, x=1

=>x^3-7x^2+15x-9

=>(1)^3-7(1)^2+15(1)-9

=>1-7+15-9

=>16-16

=>0

Therefore, (x-1) is a factor of x^3-7x^2+15x -9. =>x^3-7x^2+15x-9=(x-1)(x^2-6x+9)

=>(x-1)(x^2-3x-3x+9)

=>(x-1)(x(x-3)-3(x-3))

=>(x-1)(x-3)(x-3)

=>(x-1)(x-3)^2

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Show that (x-1) is a factor of x^3-7x^2+15x -9

Show that (x-1) is a factor of x^3-7x^2+15x -9
Hence
factorise