Can $y= x^2+7x-30$ be factored? If so what are the factors ?

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Answer 1 · 2016-01-04T00:16:39

$x^2+7x-30 = (x+10)(x-3)$

Find a pair of factors of $30$ which multiply to give $30$ and differ by $7$ . The pair $10$ , $3$ works, hence:

$x^2+7x-30 = (x+10)(x-3)$

$x^2+7x-30$ is in the form $ax^2+bx+c$ with $a=1$ , $b=7$ and $c=-30$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = 7^2-(4xx1xx-30) = 49+120 = 169 = 13^2$

Since this is a perfect square, the quadratic has two linear factors with rational coefficients.

Rather than search for a suitable pair of factors of $30$ we could use the quadratic formula to find the zeros of $x^2+7x-30$ and hence its factors:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)$

$=(-7+-13)/2$

That is $x=-10$ or $x=3$

Hence factors $(x+10)$ and $(x-3)$

Can y= x^2+7x-30 y= x^2+7x-30 be factored? If so what are the factors ?