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Mathematics → Precalculus → Quadratic Equations → Factor Quadratic Equations

Instructions for problem set  Factor each quadratic equation

Questions


\(y = x^2 + 4 x + 4\)

\(y = 4 x^2 -10 x + 4\)

\(y = x^2 -3 x -4\)

\(y = x^2 + 10 x + 24\)

\(y = 2 x^2 -7 x + 6\)

\(y = x^2 + 8 x + 15\)

\(y = x^2 + 7 x + 12\)

\(y = 2 x^2 + 5 x -3\)

\(y = 4 x^2 -5 x + 1\)

\(y = 2 x^2 + x -6\)

\(y = x^2 -x -6\)

\(y = 3 x^2 -5 x -2\)

\(y = 4 x^2 + 2 x -2\)

\(y = x^2 + x -6\)

\(y = 4 x^2 + 14 x + 6\)

\(y = x^2 + 3 x + 2\)

\(y = x^2 -7 x + 10\)

\(y = x^2 -x -20\)

\(y = 2 x^2 -x\)

\(y = x^2 -2 x -3\)

\(y = x^2 -25\)

\(y = 2 x^2 -6 x\)

\(y = x^2 + 4 x -5\)

\(y = 4 x^2 + 9 x + 2\)

\(y = x^2 + x -12\)

\(y = x^2 + 12 x + 36\)

\(y = x^2 -12 x + 36\)

\(y = x^2 + 8 x + 16\)

Answers


\(y = x^2 + 4 x + 4 \) ⇒
\((x + 2)(x + 2)\)
\(y = 4 x^2 -10 x + 4 \) ⇒
\(2(2 x -1)(x -2)\)
\(y = x^2 -3 x -4 \) ⇒
\((x -4)(x + 1)\)
\(y = x^2 + 10 x + 24 \) ⇒
\((x + 4)(x + 6)\)

\(y = 2 x^2 -7 x + 6 \) ⇒
\((2 x -3)(x -2)\)
\(y = x^2 + 8 x + 15 \) ⇒
\((x + 5)(x + 3)\)
\(y = x^2 + 7 x + 12 \) ⇒
\((x + 3)(x + 4)\)
\(y = 2 x^2 + 5 x -3 \) ⇒
\((2 x -1)(x + 3)\)

\(y = 4 x^2 -5 x + 1 \) ⇒
\((4 x -1)(x -1)\)
\(y = 2 x^2 + x -6 \) ⇒
\((2 x -3)(x + 2)\)
\(y = x^2 -x -6 \) ⇒
\((x -3)(x + 2)\)
\(y = 3 x^2 -5 x -2 \) ⇒
\((3 x + 1)(x -2)\)

\(y = 4 x^2 + 2 x -2 \) ⇒
\(2(2 x -1)(x + 1)\)
\(y = x^2 + x -6 \) ⇒
\((x -2)(x + 3)\)
\(y = 4 x^2 + 14 x + 6 \) ⇒
\(2(2 x + 1)(x + 3)\)
\(y = x^2 + 3 x + 2 \) ⇒
\((x + 1)(x + 2)\)

\(y = x^2 -7 x + 10 \) ⇒
\((x -5)(x -2)\)
\(y = x^2 -x -20 \) ⇒
\((x -5)(x + 4)\)
\(y = 2 x^2 -x \) ⇒
\((2 x -1)(x)\)
\(y = x^2 -2 x -3 \) ⇒
\((x -3)(x + 1)\)

\(y = x^2 -25 \) ⇒
\((x + 5)(x -5)\)
\(y = 2 x^2 -6 x \) ⇒
\(2(x)(x -3)\)
\(y = x^2 + 4 x -5 \) ⇒
\((x + 5)(x -1)\)
\(y = 4 x^2 + 9 x + 2 \) ⇒
\((4 x + 1)(x + 2)\)

\(y = x^2 + x -12 \) ⇒
\((x -3)(x + 4)\)
\(y = x^2 + 12 x + 36 \) ⇒
\((x + 6)(x + 6)\)
\(y = x^2 -12 x + 36 \) ⇒
\((x -6)(x -6)\)
\(y = x^2 + 8 x + 16 \) ⇒
\((x + 4)(x + 4)\)
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