Question

Me pueden ayudar porfavor estos ejercicios son de producto

Answer 1

Respuesta:

$\left(x-1\right)\left(x+4\right)\\\\=xx+x\cdot \:4+\left(-1\right)x+\left(-1\right)\cdot \:4\\\\=xx+4x-1\cdot \:x-1\cdot \:4\\=x^2+3x-4$

$\left(x-2\right)\left(x-3\right)\\\\=xx+x\left(-3\right)+\left(-2\right)x+\left(-2\right)\left(-3\right)\\\\=xx-3x-2x+2\cdot \:3\\\\=x^2-5x+6$

$\left(x+6\right)\left(x-2\right)\\\\=xx+x\left(-2\right)+6x+6\left(-2\right)\\\\=xx-2x+6x-6\cdot \:2\\\\=x^2+4x-12$

$\left(x+3\right)\left(x-1\right)\\\\=xx+x\left(-1\right)+3x+3\left(-1\right)\\\\=xx-1\cdot \:x+3x-3\cdot \:1\\\\=x^2+2x-3$

$\left(x+8\right)\left(x-6\right)\\\\=xx+x\left(-6\right)+8x+8\left(-6\right)\\\\=xx-6x+8x-8\cdot \:6\\\\=x^2+2x-48$

$\left(a-5\right)\left(a-1\right)\\\\=aa+a\left(-1\right)+\left(-5\right)a+\left(-5\right)\left(-1\right)\\\\=aa-1\cdot \:a-5a+5\cdot \:1\\\\=a^2-6a+5$

$\left(y+10\right)\left(y+12\right)\\\\=yy+y\cdot \:12+10y+10\cdot \:12\\\\=yy+12y+10y+10\cdot \:12\\\\=y^2+22y+120$

Answer 2

Respuesta:

espero te ayude :)

Explicación paso a paso:

$1)(x+1)(x+2)= x^{2} +2x+3x+2=x^{2} +5x+2$

$2) (x+3)(x+4)=x^{2} +4x+3x+12=x^{2} +7x +12$

$3) (x-1)(x + 4)=x^{2} +4x -x -4 = x^{2} +3x -4$

$4)( x-2)(x-3)= x^{2} -3x -2x +6 = x^{2} -5x+6$

$5) (x+6)(x-2)=x^{2} -2x+6x-12 = x^{2} +4x-12$

$6) (a+3)(a-1)=a^{2} -a + 3a - 3= a^{2} +2a-3$

$7) (a+8)(a-6)=a^{2} -6a+8a-48 = a^{2} +2a -48$

$8) (a-5)(a-1)=a^{2} -a -5a+5=a^{2}-6a+5$

$9) (y+10)(y+12)=y^{2} +12y +10y+120=y^{2} +22y+120$

$10) (b+8)(b-12)=b^{2} -12b+8b-96=b^{2} -4b -96$

$11) (2+2y)(y+3y)= 2y + 6y +2y^{2} +6y^{2} =8y + 8y^{2}$

$12) (x+4y)(x-2y)=x^{2} -2yx + 4yx -8y^{2} = x^{2} +2yx -8y^{2}$

$13) (a-3b)(a+5b)=a^{2} +5ba -3ba -15b^{2} =a^{2} +5ba -15b^{2}$

$14)(ab + 3)(ab-4)=ab^{2} -4ab +3ab -12=ab^{2} -ab -12$

$15)(ab+2c)(ab-4c)=ab^{2} -4cab +2cab -8c^{2} =ab^{2} -2cab -8c^{2}$