Find a and b such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$.

Question

Find $a$ and $b$ such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$.

That is what the question states, I think you should factor

Any Ideas on how to begin?

Answer 1

Expanding $(x+a)(x+b)$ gives $x^2+(a+b)x+ab$. So you want $a+b=8$ and $ab=7$. Can you see any values of $a,b$ that satisfy these two equations?

Answer 2

By factorization, $$x^2+8x+7 \equiv (x+7)(x+1) \equiv (x+a)(x+b)$$

$$\therefore \quad (a,b)=(1,7) \quad \text{or} \quad (7,1)$$

Answer 3

This factorisation is equivalent to finding roots $-a$, $-b$ of $x^2+8x+7=0$. Now the rational root theorem says that rational roots, if any, are integers, and these integers are divisors of the constant term. Thus there are only $4$ possibilities: $\pm 1,\pm7$. Further, one sees the roots have to be negative, because of the signs of the coefficients.

Indeed, $-1$ and $-7$ are roots, whence the factorisation $$x^2+8x+7=(x+1)(x+7).$$

Answer 4

Use $$x=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a},\quad \textrm{where }ax^2+bx+c=0$$

which gives $x=-1 \textrm{ or } -7$ , so factorize will be $(x+1)(x+7)$.