Factorise the following polynomials.
(a) 6p(p – 3) + 1 (p – 3)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
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Answered by
1
Answer:
(a)6p²+17p-3 (b)7(3y−5z)2(6y−10z+1)
Step-by-step explanation:
(a) 6p(p−3)+1(p−3)
=6²−18+1(−3)
=6p²-18p+p−3
6²−17−3
(b) 14(3y−5z)
3
+7(3y−5z)
2
=7[2(3y−5z)
3
+(3y−5z)
2
]
14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2[2(3y - 5z) + 1]14(3y−5z)
3
+7(3y−5z)
2
=7(3y−5z)
2
[2(3y−5z)+1]
14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2[2(3y) - 2(5z) + 1]14(3y−5z)
3
+7(3y−5z)
2
=7(3y−5z)
2
[2(3y)−2(5z)+1]
14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2(6y - 10z + 1)14(3y−5z)
3
+7(3y−5z)
2
=7(3y−5z)
2
(6y−10z+1)
Therefore, 14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2(6y - 10z + 1)14(3y−5z)
3
+7(3y−5z)
2
=7(3y−5z)
2
(6y−10z+1)
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