# Find the value of a and b in x^{2} - 16x + a = (x + b)^{2} .

The coefficients of two polynomials can be equal and identical for each different type of term.

## Answer: The value of a is 64 and the value of b is 8 for x^{2} - 16x + a = (x + b)^{2}.

Let's find a and b.

**Explanation:**

Given, x^{2} - 16x + a = (x + b)^{2}

By expanding the RHS using (a + b)^{2 }= a^{2 }+ 2ab + b^{2}, we get

x^{2} - 16x + a = x^{2 }+ 2xb + b^{2}

On solving this equation, we get

- 16x + a = 2xb + b^{2 }--- (1)

By equating the coefficients of 'x' in the equation (1), we get

- 16 = 2 b

b = - 16 / 2

b = - 8

By equating the values of constants from in the equation (1), we get

a = b^{2 }

a = (- 8)^{2 }(Using value of b)

a = 64

### Thus, the value of a is 64 and the value of b is 8 for x^{2} - 16x + a = (x + b)^{2} .

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