Question

drag the tiles to the correct boxes to complete the pairs
match each polynomial function with one of its factors.
$f(x) = x^3 - 3x^2 - 13x + 15$
$f(x) = x^4 + 3x^3 - 8x^2 + 5x - 25$
$f(x) = x^3 - 2x^2 - x + 2$
$f(x) = -x^3 + 13x - 12$
$x - 2$
$x + 3$
$x + 4$
$x + 5$

Explanation:

Step1: Use Factor Theorem: $f(a)=0$ means $x-a$ is factor.

For $x-2$, test $f(2)$:

• $f(2)=2^3-3(2)^2-13(2)+15=8-12-26+15=-15≠0$
• $f(2)=2^4+3(2)^3-8(2)^2+5(2)-25=16+24-32+10-25=-7≠0$
• $f(2)=2^3-2(2)^2-2+2=8-8-2+2=0$
• $f(2)=2^3+13(2)-12=8+26-12=22≠0$

Step2: Test $x+3$ (i.e., $x=-3$)

$f(-3)=(-3)^3-3(-3)^2-13(-3)+15=-27-27+39+15=0$
Other polynomials:

• $f(-3)=(-3)^4+3(-3)^3-8(-3)^2+5(-3)-25=81-81-72-15-25=-112≠0$
• $f(-3)=(-3)^3-2(-3)^2-(-3)+2=-27-18+3+2=-40≠0$
• $f(-3)=(-3)^3+13(-3)-12=-27-39-12=-78≠0$

Step3: Test $x+4$ (i.e., $x=-4$)

$f(-4)=(-4)^3+13(-4)-12=-64-52-12=-128≠0$
Wait, correct: $f(-4)=(-4)^3+13(-4)-12$ no, use $f(-4)=(-4)^3+13(-4)-12$ no, test $f(-4)=(-4)^3+13(-4)-12$ no, correct: $f(-4)=(-4)^3 +13(-4)-12=-64-52-12=-128≠0$. Wait, $f(-4)=(-4)^3+13(-4)-12$ no, test $f(-4)= -(-4)^3 +13(-4)-12$? No, the polynomial is $f(x)=x^3+13x-12$. $f(-4)=(-4)^3+13(-4)-12=-64-52-12=-128≠0$. Wait, $f(1)=1+13-12=2≠0$, $f(-1)=-1-13-12=-26≠0$, $f(3)=27+39-12=54≠0$, $f(-4)$ no. Wait, $f(-4)$ for $f(x)=x^3-3x^2-13x+15$: $-64-36+52+15=-33≠0$. Wait, $f(-4)$ for $f(x)=x^4+3x^3-8x^2+5x-25$: $256-192-128-20-25=-109≠0$. Wait, $f(-4)$ for $f(x)=x^3-2x^2-x+2$: $-64-32+4+2=-90≠0$. Wait, I made a mistake: $x+4$ is $x=-4$, test $f(x)=x^3+13x-12$: $(-4)^3 +13*(-4)-12=-64-52-12=-128≠0$. Wait, test $f(1)=1+13-12=2≠0$, $f(3)=27+39-12=54≠0$, $f(-3)=-27-39-12=-78≠0$, $f(4)=64+52-12=104≠0$. Wait, no, the polynomial is $f(x)=x^3+13x-12$? No, it's $f(x)=x^3 +13x -12$. Wait, $x+4$: let's do polynomial division for $x^3+13x-12$ by $x+4$:
$x^3+0x^2+13x-12 = (x+4)(x^2-4x+29) - 128$, remainder -128. Wait, test $x=1$ for $f(x)=x^3-3x^2-13x+15$: $1-3-13+15=0$, so $x-1$ is factor, not $x+4$. Wait, $x+4$: test $f(-4)$ for $f(x)=x^3+13x-12$ no. Wait, $f(x)=x^3+13x-12$: $f(1)=1+13-12=2$, $f(-1)=-1-13-12=-26$, $f(2)=8+26-12=22$, $f(-2)=-8-26-12=-46$, $f(3)=27+39-12=54$, $f(-3)=-27-39-12=-78$, $f(4)=64+52-12=104$, $f(-4)=-64-52-12=-128$. Wait, maybe I misread the polynomial: it's $f(x)=-x^3+13x-12$? Oh yes! $f(x)=-x^3+13x-12$. Then $f(-4)= -(-64)+13*(-4)-12=64-52-12=0$. Correct!

Step4: Test $x+5$ (i.e., $x=-5$)

$f(-5)=(-5)^4+3(-5)^3-8(-5)^2+5(-5)-25=625-375-200-25-25=0$
Other polynomials:

• $f(-5)=(-5)^3-3(-5)^2-13(-5)+15=-125-75+65+15=-120≠0$
• $f(-5)=(-5)^3-2(-5)^2-(-5)+2=-125-50+5+2=-168≠0$
• $f(-5)=-(-5)^3+13(-5)-12=125-65-12=48≠0$

Answer:

$x-2
ightarrow f(x)=x^3-2x^2-x+2$
$x+3
ightarrow f(x)=x^3-3x^2-13x+15$
$x+4
ightarrow f(x)=x^3+13x-12$ (corrected to $-x^3+13x-12$ which matches)
$x+5
ightarrow f(x)=x^4+3x^3-8x^2+5x-25$