Problem 24 - Introductory Algebra for College Students

Question

Factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$y^{2}-10 y+21$$

Step-by-Step Solution

Answer

The factorization of the trinomial \(y^{2}-10 y+21\) is \( (y-7)(y-3) \).

Step 1: Identifying the Factors

The trinomial is \(y^{2}-10 y+21\). We need to find two numbers that multiply to give +21 (the third term in the trinomial) and add up to give -10 (the coefficient of the second term). The numbers that satisfy these conditions are -7 and -3, because -7 multiplied by -3 gives +21 and -7 plus -3 adds up to -10.

Step 2: Factoring the Trinomial

Now that we have identified the factors, the trinomial can be factored by substituting these values into two binomials. So \( y^{2}-10y+21 \) can be factored into \( (y-7)(y-3) \). Therefore, \( y^{2}-10y+21 = (y-7)(y-3) \) is the factorization of the given trinomial.

Step 3: Checking the Factorization using FOIL

To verify if the factorization is correct, we can use the FOIL (First, Outer, Inner, Last) method. Multiply the first term in each binomial to get the first term of the trinomial, \(y*y\) which gives \(y^2\). Then multiply the outer terms in the binomials to get \(-7y\), the inner terms to get \(-3y\). Adding these gives \(-10y\). Lastly, multiply the last terms in each binomial to get \(21\). This shows that \( y^{2}-10y+21 =(y-7)(y-3)\), confirming that the factorization is correct.