Please help me solve this before 12am today.

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robsantana11

18.12.2022

Mathematics

The answers are given below.

Step-by-step explanation: We are given to factorise some quadratic expressions.

We will be using the following factorisation formula:

x^2+ax+bx+ab=(x+a)(x+b).

The factorisation is as follows:

We have

$(1)~3ab-5ax+9b-15x\\\\=a(3b-5x)+3(3b-5x)\\\\=(3b-5x)(a+3).$

Thus, option (B) (3b-5x)(a+3) is correct.

$(2)~5x(x+3)+6(x+3)\\\\=(x+3)(5x+6).$

Thus, option (B) (x+3)(5x+6) is correct.

$(3)~2cb+7cx+2b+7x\\\\=c(2b+7x)+1(2b+7x)\\\\=(2b+7x)(c+1).$

Thus, option (B) (2b+7x)(c+1) is correct

$(4)~21x^3+35x^2+9x+15\\\\=7x^2(3x+5)+3(3x+5)\\\\=(3x+5)(7x^2+3).$

Thus, option (D) (3x+5)(7x^2+3) is correct.

$(5)~9xy+36x-10y-40\\\\=9x(y+4)-10(y+4)=(y+4)(9x-10).$

Thus, option (C) (y+4)(9x-10) is the correct option.

$(6)~14bx^2-7x^3-4b+2x\\\\=7x^2(2b-x)-2(2b-x)\\\\=(2b-x)(7x^2-2).$

Thus, option (B) (2b-x)(7x^2-2) is the correct option.

$(7)~4xy+9x+24y+54\\\\=x(4y+9)+6(4y+9)\\\\=(4y+9)(x+6).$

Thus, option (D) (4y+9)(x+6) is correct.

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