Factorise : [4 MARKS]
1)x3−2x2−x+2
2)2y3+y2−2y−1
Factorise : [4 MARKS]
1)x3−2x2−x+2
2)2y3+y2−2y−1
Concept: 1 Mark
Factorisation: 1.5 Marks each
1)Let p(x)=x3−2x2−x+2
p(1)=13−2(1)2−1+2
=1−2−1+2=0 [Using hit and trial method]
By Factor theorem, we can say that (x−1) 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)=x3−2x2−x+2
=x3−x2−x2+x−2x+2
=x2(x−1)−x(x−1)−2(x−1)
=(x−1)(x2−x−2)
=(x−1)(x−2)(x+1)
2)p(y)=2y3+y2−2y−1
p(1)=2(1)3+(1)2−2(1)−1=2+1−2−1=0[Using hit and trial method]
Therefore, by factor theorem, we can say that (y−1) 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+y2−2y−1
=2y3+2y2−y2−y−y−1
=2y2(y+1)−y(y+1)−1(y+1)
=(y+1)(2y2−y−1)
=(y+1)(y−1)(2y+1)
Concept: 1 Mark
Factorisation: 1.5 Marks each
1)Let p(x)=x3−2x2−x+2
p(1)=13−2(1)2−1+2
=1−2−1+2=0 [Using hit and trial method]
By Factor theorem, we can say that (x−1) 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)=x3−2x2−x+2
=x3−x2−x2+x−2x+2
=x2(x−1)−x(x−1)−2(x−1)
=(x−1)(x2−x−2)
=(x−1)(x−2)(x+1)
2)p(y)=2y3+y2−2y−1
p(1)=2(1)3+(1)2−2(1)−1=2+1−2−1=0[Using hit and trial method]
Therefore, by factor theorem, we can say that (y−1) 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+y2−2y−1
=2y3+2y2−y2−y−y−1
=2y2(y+1)−y(y+1)−1(y+1)
=(y+1)(2y2−y−1)
=(y+1)(y−1)(2y+1)
