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प्रश्न
Factorise the following by taking out the common factor
3y3 – 48y
योग
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उत्तर
3y2 – 48y = 3 × y × y2 – 3 × 16 × y
Taking out 3 × y
= 3y(y2 – 16) = 3y(y2 – 42)
Comparing y2 – 42 with a2 – b2
a = y, b = 4
a2 – b2 = (a + b)(a – b)
y2 – 42 = (y + 4)(y – 4)
∴ 3y(y2 – 16) = 3y(y + 4)(y – 4)
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क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Algebra - Exercise 3.4 [पृष्ठ ९६]
