Question

factor the following trinomial.
b² − 3b − 10
(b − ?)(b + )

Explanation:

Step 1: Identify the form of factoring

We need to factor the trinomial \(b^2 - 3b - 10\) into the form \((b - m)(b + n)\), where \(m\) and \(n\) are integers. When we expand \((b - m)(b + n)\), we get \(b^2+(n - m)b - mn\). Comparing this with \(b^2 - 3b - 10\), we have two equations: \(n - m=- 3\) and \(mn = 10\).

Step 2: Find integers m and n

We need to find two integers \(m\) and \(n\) such that their product is \(10\) and their difference (\(n - m\)) is \(-3\). Let's list the factor pairs of \(10\): \((1,10)\), \((2,5)\), \((-1,-10)\), \((-2,-5)\).

• For the pair \((2,5)\): If \(m = 5\) and \(n=-2\) (since we need \(n - m=-3\), let's check \(n - m=-2 - 5=-7

eq - 3\)). Wait, if we take \(m = 5\) and \(n = 2\), then \(n - m=2 - 5=-3\) and \(mn=5\times2 = 10\). Perfect!

Step 3: Substitute m and n into the factors

So, \(m = 5\) and \(n=-2\). So the factored form is \((b - 5)(b+(- 2))=(b - 5)(b - 2)\)? Wait, no, wait. Wait the form is \((b - m)(b + n)\). Wait, let's re - check. Wait, when we have \((b - m)(b + n)=b^2+(n - m)b - mn\). We have \(b^2-3b - 10\), so \(n - m=-3\) and \(mn = 10\). Let's solve for \(m\) and \(n\) from \(n=m - 3\). Substitute into \(mn = 10\): \(m(m - 3)=10\), \(m^2-3m - 10 = 0\). Factoring this quadratic: \(m^2-3m - 10=(m - 5)(m + 2)=0\). So \(m = 5\) or \(m=-2\). If \(m = 5\), then \(n=5 - 3 = 2\)? Wait no, wait \(n - m=-3\) so \(n=m - 3\). If \(m = 5\), \(n=5-3 = 2\)? But then \(mn=5\times2 = 10\). But then \((b - m)(b + n)=(b - 5)(b + 2)\). Wait, but when we expand \((b - 5)(b + 2)=b^2+2b-5b - 10=b^2-3b - 10\), which is correct. Wait, the original form was \((b - [?])(b + [\square])\). So if we have \((b - 5)(b + 2)\), then the first blank is \(5\) and the second is \(2\). Wait, maybe I made a mistake in the sign earlier. Let's re - do the expansion. \((b - m)(b + n)=b^2+nb - mb - mn=b^2+(n - m)b - mn\). We have \(b^2-3b - 10\), so \(n - m=-3\) and \(-mn=-10\) (since the constant term is \(-10\)), so \(mn = 10\). So we need two numbers that multiply to \(10\) and add up to \(-3\) (wait, no: \(n - m=-3\), so \(n=m - 3\). And \(mn = 10\). So substituting \(n=m - 3\) into \(mn = 10\) gives \(m(m - 3)=10\), \(m^2-3m - 10 = 0\). Factoring: \(m^2-3m - 10=(m - 5)(m + 2)=0\), so \(m = 5\) or \(m=-2\). If \(m = 5\), then \(n=5 - 3 = 2\). Then \((b - 5)(b + 2)=b^2+2b-5b - 10=b^2-3b - 10\), which is correct. So the first blank (the one in \(b - [?]\)) is \(5\) and the second blank (in \(b + [\square]\)) is \(2\).

Answer:

The first blank (the green one) is \(5\) and the second blank (the white one) is \(2\). So the factored form is \((b - 5)(b + 2)\), so the value in the green box is \(5\) and the value in the white box is \(2\).