1(a.) Factorize x^2+3x-28 = (x+7)(x-4) I know how to do 1(b). Hence factorize (y-4)^2-4(y^2+3y-28) = -(y-4)(3y+32) I want the steps.plz!!!?
回答 (2)
2015-10-24 4:04 am
1(a.) Factorize x^2+3x－28
x^2+3x－28
=(x^2+7x)+(－4x－28)
=x(x+7)－4(x+7)
=(x－4)(x+7)
(b). Hence factorize (y－4)^2－4(y^2+3y－28) = －(y－4)(3y+32)
(y－4)^2－4(y^2+3y－28)
=(y－4)^2－4(y－4)(y+7)
=(y－4)^2－(y－4)(4y+28)
=(y－4)[(y－4)－(4y+28)]
=(y－4)(－3y－32)
=－(y－4)(3y+32)
x^2+3x－28
=(x^2+7x)+(－4x－28)
=x(x+7)－4(x+7)
=(x－4)(x+7)
(b). Hence factorize (y－4)^2－4(y^2+3y－28) = －(y－4)(3y+32)
(y－4)^2－4(y^2+3y－28)
=(y－4)^2－4(y－4)(y+7)
=(y－4)^2－(y－4)(4y+28)
=(y－4)[(y－4)－(4y+28)]
=(y－4)(－3y－32)
=－(y－4)(3y+32)
2015-10-23 3:26 pm
1.十字交乘
x^2+3x-28=(x+7)(x-4)
2.
(y-4)^2-4(y^2+3y-28)=(y-4)^2-4[(y-4)^2+11y-44]=(y-4)^2-4[(y-4)^2+11(y-4)]=(y-4)^2-4(y-4)[(y-4)+11]
=(y-4)^2-4(y-4)(y+7)=(y-4)[(y-4)-4(y+7)]=(y-4)(-3y-32)= -(y-4)(3y+32)
x^2+3x-28=(x+7)(x-4)
2.
(y-4)^2-4(y^2+3y-28)=(y-4)^2-4[(y-4)^2+11y-44]=(y-4)^2-4[(y-4)^2+11(y-4)]=(y-4)^2-4(y-4)[(y-4)+11]
=(y-4)^2-4(y-4)(y+7)=(y-4)[(y-4)-4(y+7)]=(y-4)(-3y-32)= -(y-4)(3y+32)
收錄日期: 2021-04-18 14:01:22
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