# ax^2 + bx + c = ( x-2)(x+5) find the values of a, b and c:

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ax^2 + bx + c = ( x-2)(x+5)

First we will expand the brackets of the right sides:

==> ax^2 + bx + c = x^2 +5x - 2x - 10

==> ax^2 + bx + c = x^2 + 3x - 10

Now we will move all terms to the left side:

==> ax62 + bx + x - x^2 - 3x + 10 = 0

Now rearrange terms:

==> ax^2 - x^2 + bx-3x + c +10 = 0

Now combine like terms:

==> (a-1)x^2 + (b-3)x + (c+10) = 0

Since the Right side is 0: then each terms will equal o:

==> a-1 = 0 ==> a = 1

==> b-3 = 0 ==> b= 3

==> (c+10) = 0 ==> c = -10

Then the answer is:

**a = 1, b = 3 and c = -10**

**x^2 + 3x - 10 = (x-2)(x+5)**

To find a b, c in ax^2+bx+c = (x-2)(x-b)

We edit ax^2+bx+x to be ax^2+bx+c.

Solution:

ax^2+bx+c= (x-2)(x+5)..(1)

If the above is an identity, then it must be true for all values of x.

So put x= 0 ineq (1)

a0^2+b*0 +c = (0-2)(0+5).

So c = -2*5 = -10

Put x = 2 in eq ax^2+bx-10 = (x-2)(x+5)

a*4+b*2 -10 = 0*(0+5) = 0. So (4a+2b -10) = 0

2a+b = 5....(2)

Put x = -5 in ax^2+b -10 = 0 .Then a*25-5b-10 = = = 0.

Or 5a - b= 2....(3).

(2)+(3) gives: 7a = 7, or a = 1.

Put a = 1 in (2), then b = 5-2a = 5-2 = 3.

Therefore a = 1, b= 3 and c = -10.