Please solve this question. right answer will be marked brainlist n i will follo you.
Answers
Answer:
Correct option is
A
(3y−1)(5y−1)
Let p(y)=15y
2
−8y+1=15×{y
2
−
15
8
y+
15
1
}
=15×q(y)
where q(y)=y
2
−
15
8
y+
15
1
we have made the coefficient of the leading term of the polynomial q(y) equal to 1. The constant term of the quadratic polynomial q(y) is
15
1
. Some of the factors of the number
15
1
are ±1,±
3
1
,±
5
1
,±
15
1
.
we find that
q(
3
1
)=(
3
1
)
2
−
15
8
(
3
1
)+
15
1
=
9
1
−
45
8
+
15
1
=
45
5−8+3
=0
and q(
5
1
)=(
5
1
)
2
−
15
8
(
5
1
)+
15
1
=
25
1
−
75
8
+
15
1
=
75
3−8+5
=0
Thus,
3
1
and
5
1
are zeros of q(y). It implies that
(y−
3
1
) and (y−
5
1
) are two
factors of the polynomial q(y)=y
2
−
15
8
y+
15
1
.
(By factor theorem).
i.e., y
2
−
15
8
+
15
1
=(y−
3
1
)(y−
5
1
)
⇒15y
2
−8y+1=15×{(y−
3
1
)(y−
5
1
)}
=15×{
2
(3y−1)
×
5
(5y−1)
}
=(3y−1)(5y−1)
Therefore, the polynomial 15y
2
−8y+1 factorised into two linear factors as (3y-1)(5y-1).
Answer:
Answer in the image !!
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