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Question

Factorise : [4 MARKS]
1)x32x2x+2
2)2y3+y22y1

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Solution

Concept: 1 Mark
Factorisation: 1.5 Marks each

1)Let p(x)=x32x2x+2

p(1)=132(1)21+2
=121+2=0 [Using hit and trial method]

By Factor theorem, we can say that (x1) is the factor of p(x)

We can also find other factors of p(x) by doing long division but we will just split the terms to take (x − 1) common from p(x).

p(x)=x32x2x+2

=x3x2x2+x2x+2

=x2(x1)x(x1)2(x1)

=(x1)(x2x2)

=(x1)(x2)(x+1)

2)p(y)=2y3+y22y1

p(1)=2(1)3+(1)22(1)1=2+121=0[Using hit and trial method]

Therefore, by factor theorem, we can say that (y1) is a factor of p(y).

Now, we will split the terms accordingly to take (y−1) common from p(y). We can also do long division to find other factors but it may take more time.

p(y)=2y3+y22y1

=2y3+2y2y2yy1

=2y2(y+1)y(y+1)1(y+1)

=(y+1)(2y2y1)

=(y+1)(y1)(2y+1)


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