Question

If (5x− 4)(3x+ 7) = 15x²-ax-b,then find the value of a and b

Answer 1

Answer:

$a=-23\\\\b=28$

Step-by-step explanation:

Let's start by using distributive multiplication:

$(a\pm b)(c\pm d)=ac\pm ad \pm bc \pm bd$

So:

$(5x-4)(3x+7)=15x^2+35x-12x-28$

Grouping like terms:

$15x^2 +(35x-12x)-28\\\\15x^2+23x-28$

Now, $15x^2+23x-28$ is equal to:

$15x^2-ax-b$

In this sense:

$15x^2+23x-28=15x^2-ax-b$

In order to satisfied the equality:

$23x=-ax\hspace{10}(1)\\\\and\\\\-28=-b\hspace{10}(2)$

Hence, from (1), let's solve for a:

$23x=-ax\\\\-a=\frac{23x}{x} \\\\-a=23\\\\a=-23$

And from (2), let's solve for b:

$-28=-b\\\\b=28$

Let's verify the result evaluating the values of a and b into the original equation:

$(5x-4)(3x+7)=15x^2+23x-28=15x^2 -ax -b\\\\(5x-4)(3x+7)=15x^2+23x-28=15x^2 -(-23)x -(28)\\\\(5x-4)(3x+7)=15x^2+23x-28=15x^2 +23x -28$

As you can see, the values satisfy the equation, therefore, we can conclude they are correct.