Maths question!?

2016-06-22 10:25 am
how to factorise these WITH working out please:

1.
2ap-a^2+6bp-3ab

2.
2m^2n-nm+m^3-2n^2

3.
p^2q^2+p^r-rq^2-r^2

4.
p^4-pq+q^4-p^3q^3

Thanks a lot :)

回答 (3)

2016-06-22 10:47 am
✔ 最佳答案
= 2ap - a² + 6bp - 3ab

= 2ap + 6bp - a² - 3ab

= (2ap + 6bp) - (a² + 3ab)

= 2p.(a + 3b) - a.(a + 3b)

= (a + 3b).(2p - a)


= 2m²n - nm + m³ - 2n²

= 2m²n - 2n² - nm + m³

= (2m²n - 2n²) - (nm - m³)

= 2n.(m² - n) - m.(n - m²)

= 2n.(m² - n) + m.(m² - n)

= (m² - n).(2n + m)


= p²q² + p² - rq² - r

= (p²q² + p²) - (rq² + r)

= p².(q² + 1) - r.(q² + 1)

= (q² + 1).(p² - r)


= p⁴ - pq + q⁴ - p³q³

= (p⁴ - pq) + (q⁴ - p³q³)

= p.(p³ - q) + q³.(q - p³)

= p.(p³ - q) - q³.(p³ - q)

= (p³ - q).(p - q³)
2016-06-22 10:53 am
1.
2ap - a² + 6bp - 3ab
= (2ap - a²) + (6bp - 3ab)
= a(2p - a) + 3p(2p - a)
= (3p + a)(2p - a)


2.
2m²n - nm + m³ - 2n²
= m³ + 2m²n - nm - 2n²
= (m³ + 2m²n) - (nm + 2n²)
= m²(m + 2n) - n(m + 2n)
= (m² - n)(m + 2n)


3.
p²q² + p²r - rq² - r²
= (p²q² + p²r) - (rq² - r²)
= p²(q² - r) - r(q² - r)
= (p² - r)(q² - r)


4.
p⁴ - pq + q⁴ - p³q³
= p⁴ - p³q³ - pq + q⁴
= (p⁴ - p³q³) - (pq - q⁴)
= p³(p - q³) - q(p - q³)
= (p³ - q)(p - q³)
2016-06-22 1:57 pm
1.
2ap - a^2 + 6bp - 3ab
= (a + 3b)(2p - a)
2.
2m^2n - nm + m^3 - 2n^2
= (m^2 - n)(m + 2n)
3.
p^2q^2 + p^2 - (rq)^2 - r^2
= (q^2 + 1)(p^2 - r^2)
4.
p^4 - pq + q^4 - p^3q^3
= (p^3 - q)(p - q^3)


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