What are the two binomial factors for 4x^2-41x+72

To factor the quadratic expression $4x^2 - 41x + 72$, we need to find two binomial factors that multiply to give the original expression.<br /><br />The general form of a quadratic expression is $ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. In this case, $a = 4$, $b = -41$, and $c = 72$.<br /><br />To factor the quadratic expression, we need to find two numbers whose product is equal to $ac$ (which is $4 \cdot 72 = 288$) and whose sum is equal to $b$ (which is $-41$).<br /><br />Let's call the two numbers $m$ and $n$. We need to find $m$ and $n$ such that $m \cdot n = 288$ and $m + n = -41$.<br /><br />By trial and error, we can find that $m = -9$ and $n = -32$ satisfy these conditions, since $-9 \cdot -32 = 288$ and $-9 + (-32) = -41$.<br /><br />Now, we can rewrite the quadratic expression as $4x^2 - 9x - 32x + 72$ and factor by grouping:<br /><br />$4x^2 - 9x - 32x + 72 = 4x(x - \frac{9}{4}) - 8(x - \frac{9}{4}) = (4x - 8)(x - \frac{9}{4})$<br /><br />Therefore, the two binomial factors for $4x^2 - 41x + 72$ are $(4x - 8)$ and $(x - \frac{9}{4})$.