Factorise completely:
1) x² - 8x + 16
2) a² - 12a + 36
3) 3b² - 6b + 3
thankssss ;) please show step to step :P thankss
Math helpppp! Please and thankyouu!: )?
2009-04-09 4:27 pm
回答 (7)
2009-04-09 4:32 pm
✔ 最佳答案
(x - 4)²(a - 6)²
3* (b² - 2b + 1) = 3 * (b -1)²
*******
2009-04-09 11:44 pm
1) x² - 8x + 16
..=(x-4)(x-4)
=(x-4)^2 answer//
2.) a² - 12a + 36
=(a-6)(a-6)
=(a-6)^2 answer//
3,) 3b² - 6b + 3
=3(b^2-2b+1)
=3(b-1)(b-1)
=3(b-1)^2 answer//
..=(x-4)(x-4)
=(x-4)^2 answer//
2.) a² - 12a + 36
=(a-6)(a-6)
=(a-6)^2 answer//
3,) 3b² - 6b + 3
=3(b^2-2b+1)
=3(b-1)(b-1)
=3(b-1)^2 answer//
2009-04-09 11:44 pm
here u need 2 understand sth.
x2 is square of x coz i dont know how 2 do square
1) x2-8x+16=0
x2-4x-4x+16=0
x(x-4)-4(x-4)=0
(combining x and -4 & taking only one bracket from (x-4) we get)
(x-4)(x-4)=0
to get product of these to b 0 one of this bracket shud b equal to 0 therefor
x-4=0
x=4
u should factorise middle term in such a manner so that its product is = product of first no. & last no. and its sum sud b equal 2 its middle term. this method is also known as middle term spliting
x2 is square of x coz i dont know how 2 do square
1) x2-8x+16=0
x2-4x-4x+16=0
x(x-4)-4(x-4)=0
(combining x and -4 & taking only one bracket from (x-4) we get)
(x-4)(x-4)=0
to get product of these to b 0 one of this bracket shud b equal to 0 therefor
x-4=0
x=4
u should factorise middle term in such a manner so that its product is = product of first no. & last no. and its sum sud b equal 2 its middle term. this method is also known as middle term spliting
2009-04-09 11:37 pm
1) (x-4)^2
2) (a-6)^2
3) 3(b-1)^2
That's it Maths rock
2) (a-6)^2
3) 3(b-1)^2
That's it Maths rock
2009-04-09 11:34 pm
1)
x^2 - 8x + 16
= x^2 - 4x - 4x + 16
= (x^2 - 4x) - (4x - 16)
= x(x - 4) - 4(x - 4)
= (x - 4)(x - 4)
= (x - 4)^2
2)
a^2 - 12a + 36
= a^2 - 6a - 6a + 36
= (a^2 - 6a) - (6a - 36)
= a(a - 6) - 6(a - 6)
= (a - 6)(a - 6)
= (a - 6)^2
3)
3b^2 - 6b + 3
= 3(b^2 - 2b + 1)
= 3(b^2 - b - b + 1)
= 3[(b^2 - b) - (b - 1)]
= 3[b(b - 1) - 1(b - 1)]
= 3(b - 1)(b - 1)
= 3(b - 1)^2
x^2 - 8x + 16
= x^2 - 4x - 4x + 16
= (x^2 - 4x) - (4x - 16)
= x(x - 4) - 4(x - 4)
= (x - 4)(x - 4)
= (x - 4)^2
2)
a^2 - 12a + 36
= a^2 - 6a - 6a + 36
= (a^2 - 6a) - (6a - 36)
= a(a - 6) - 6(a - 6)
= (a - 6)(a - 6)
= (a - 6)^2
3)
3b^2 - 6b + 3
= 3(b^2 - 2b + 1)
= 3(b^2 - b - b + 1)
= 3[(b^2 - b) - (b - 1)]
= 3[b(b - 1) - 1(b - 1)]
= 3(b - 1)(b - 1)
= 3(b - 1)^2
2009-04-09 11:32 pm
1) x² - 8x + 16
(x–4)² no steps, by inspection
2) a² - 12a + 36
(a–6)²
3) 3b² - 6b + 3
3(b² – 2b + 1)
3(a – 1)²
they all are very similar and use:
(x–n)² = x² –2nx + n²
.
(x–4)² no steps, by inspection
2) a² - 12a + 36
(a–6)²
3) 3b² - 6b + 3
3(b² – 2b + 1)
3(a – 1)²
they all are very similar and use:
(x–n)² = x² –2nx + n²
.
2009-04-09 11:31 pm
(x-4)^2
(a-6)^2
(3b+3)(b-1)
(a-6)^2
(3b+3)(b-1)
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