Answered

(x+5)(2x-3)-3(x^(2)+2x)+5x^(2) If the expression above is rewritten in the form ax^(2)+bx+c , where a,b and c are constants, what is the value of a ? 1 1 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 5 4 4 4 6 6 5 5 7 7 7 7 8 8 8 8 9 9 9 9 1 1 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 5 4 4 4 6 6 5 5 7 7 7 7 8 8 8 8 9 9 9 9?



Answer :

Answer:

[tex]\huge\boxed{x = 4}[/tex]

Step-by-step explanation:

Given expression:

(x + 5)(2x - 3) - 3(x² + 2x) + 5x²

Distribute

= x(2x - 3) + 5(2x - 3) - 3x² - 6x + 5x²

= 2x² - 3x + 10x - 15 - 3x² - 6x + 5x²

Combine like terms

= 2x² - 3x² + 5x² - 3x + 10x - 6x - 15

= -x² + 5x² + 7x - 6x - 15

= 4x² - 6x - 15

Comparing it with ax² + bx + c, we get:

a = 4

[tex]\rule[2]{225}{2}[/tex]

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