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completely factor the polynomial, if possible. \\25x^2 + 22x + 9\\

Question

completely factor the polynomial, if possible.

\\25x^2 + 22x + 9\\

Explanation:

Identify the trinomial coefficients

We begin with the given quadratic expression:

$$25x^2 + 22x + 9$$

This is a trinomial of the form \(ax^2 + bx + c\), where:

  • \(a = 25\)
  • \(b = 22\)
  • \(c = 9\)

Check for perfect square trinomial pattern

Using the Factoring Trinomials knowledge point, we first check if this fits a perfect square trinomial pattern:

$$(px + q)^2 = p^2x^2 + 2pqx + q^2$$

Here, the first and last terms are perfect squares:

  • \(p^2 = 25 \implies p = 5\)
  • \(q^2 = 9 \implies q = 3\)

The middle term for a perfect square trinomial would be:

$$2pqx = 2(5)(3)x = 30x$$

Since our middle term is \(22x\) and not \(30x\), this expression is not a perfect square trinomial.

Analyze factoring by grouping

Next, we look for two integers whose product is \(ac\) and whose sum is \(b\):

  • Product: \(ac = 25 \times 9 = 225\)
  • Sum: \(b = 22\)

We list the factor pairs of \(225\) to see if any pair sums to \(22\):

  • \(1 \times 225 \implies \text{sum} = 226\)
  • \(3 \times 75 \implies \text{sum} = 78\)
  • \(5 \times 45 \implies \text{sum} = 50\)
  • \(9 \times 25 \implies \text{sum} = 34\)
  • \(15 \times 15 \implies \text{sum} = 30\)

None of the factor pairs of \(225\) sum to \(22\).

Determine factorability

Since there are no integers that multiply to \(225\) and sum to \(22\), the polynomial cannot be factored over the integers. Therefore, the polynomial is prime (not factorable).

Answer:

Completely factor the polynomial, if possible.
\(25x^2 + 22x + 9\)
Answer: <blank>Not factorable</blank>