Expand and simplify (1 - 2x)(5 - 3x) + (4 - x)^2.

Step-by-step solution

1. 1

   Expand the product with FOIL. Multiply each term of the first bracket by each of the second, watching the signs:

   $(1-2x)(5-3x)=5-3x-10x+6x^{2}=6x^{2}-13x+5$

   The last product $(-2x)(-3x)=+6x^{2}$ is positive because two negatives multiply to a positive.

2. 2

   Expand the square with the difference identity. Using $(a-b)^{2}=a^{2}-2ab+b^{2}$ with $a=4$, $b=x$:

   $(4-x)^{2}=16-8x+x^{2}$

   There are three terms — writing $16-x^{2}$ or $16+x^{2}$ drops the essential middle term $-8x$.

3. 3

   Add the two expansions.

   $\left(6x^{2}-13x+5\right)+\left(16-8x+x^{2}\right)$

4. 4

   Collect like terms by degree.

   $6x^{2}+x^{2}=7x^{2},\qquad-13x-8x=-21x,\qquad5+16=21$

   giving

   $7x^{2}-21x+21$

5. 5

   Verify and note the common factor. At $x=1$: the original is $(-1)(2)+9=7$, and the answer gives $7-21+21=7$ ✓. The identity holds at every integer from $-30$ to $29$ ✓. All three coefficients share a factor of $7$, so the result also equals $7\left(x^{2}-3x+3\right)$; the inner quadratic has discriminant $9-12=-3<0$, so it never factors further over the reals and the expression is always positive.